Trig Ratios & Identities - IB Mathematics Analysis and Approaches

1. Assuming $\sin(\theta) + \cos(\theta) \neq 0$, the expression $\frac{\sin^3(\theta) + \cos^3(\theta)}{\sin(\theta) + \cos(\theta)}$ simplifies to:
   Answer: $1 - \sin(\theta)\cos(\theta)$ The numerator is a sum of cubes, which can be factored as $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$. Applying this formula with $a = \sin(\theta)$ and $b = \cos(\theta)$, we get $\sin^3(\theta) + \cos^3(\theta) = (\sin(\theta) + \cos(\theta))(\sin^2(\theta) - \sin(\theta)\cos(\theta) + \cos^2(\theta))$. The expression becomes $\frac{(\sin(\theta) + \cos(\theta))(\sin^2(\theta) - \sin(\theta)\cos(\theta) + \cos^2(\theta))}{\sin(\theta) + \cos(\theta)}$. The $(\sin(\theta) + \cos(\theta))$ terms cancel. We are left with $\sin^2(\theta) - \sin(\theta)\cos(\theta) + \cos^2(\theta)$. Using the identity $\sin^2(\theta) + \cos^2(\theta) = 1$, this simplifies to $1 - \sin(\theta)\cos(\theta)$.
2. The expression $(\tan(\theta) + \cot(\theta))\sin(\theta)\cos(\theta)$ simplifies to:
   Answer: 1 First, express $\tan(\theta)$ and $\cot(\theta)$ in terms of $\sin(\theta)$ and $\cos(\theta)$. The expression becomes $\left(\frac{\sin(\theta)}{\cos(\theta)} + \frac{\cos(\theta)}{\sin(\theta)}\right)\sin(\theta)\cos(\theta)$. Combine the terms in the parenthesis with a common denominator of $\sin(\theta)\cos(\theta)$: $\left(\frac{\sin^2(\theta) + \cos^2(\theta)}{\sin(\theta)\cos(\theta)}\right)\sin(\theta)\cos(\theta)$. Using the identity $\sin^2(\theta) + \cos^2(\theta) = 1$, this simplifies to $\left(\frac{1}{\sin(\theta)\cos(\theta)}\right)\sin(\theta)\cos(\theta)$. The $\sin(\theta)\cos(\theta)$ terms cancel, leaving a result of 1.
3. If $\cos(2\alpha) = \frac{7}{25}$, where $2\alpha$ is an acute angle, find the value of $\cos^4(\alpha) - \sin^4(\alpha)$.
   Answer: $\frac{7}{25}$ The expression $\cos^4(\alpha) - \sin^4(\alpha)$ is a difference of two squares. It can be factored as $(\cos^2(\alpha) - \sin^2(\alpha))(\cos^2(\alpha) + \sin^2(\alpha))$. Using the Pythagorean identity, $\cos^2(\alpha) + \sin^2(\alpha) = 1$. The expression simplifies to $\cos^2(\alpha) - \sin^2(\alpha)$. This is the double angle identity for cosine, $\cos(2\alpha)$. Since we are given that $\cos(2\alpha) = \frac{7}{25}$, the value of the expression is $\frac{7}{25}$.
4. If $\tan(\theta) = 3$ and $\cos(\theta) < 0$, find the exact value of $\sin(\theta)$.
   Answer: $-\frac{3}{\sqrt{10}}$ Since $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} > 0$ and $\cos(\theta) < 0$, it must be that $\sin(\theta) < 0$ as well. This places $\theta$ in the third quadrant. We can construct a right-angled triangle with opposite side 3 and adjacent side 1. The hypotenuse would be $\sqrt{3^2 + 1^2} = \sqrt{10}$. From this reference triangle, $|\sin(\theta)| = \frac{3}{\sqrt{10}}$. Since $\theta$ is in the third quadrant, $\sin(\theta)$ is negative. Thus, $\sin(\theta) = -\frac{3}{\sqrt{10}}$.
5. The equation $\tan^2(x) - 2\tan(x) - 3 = 0$ holds for an acute angle $x$. Which of the following is the value of $\sin(x)$?
   Answer: $\frac{3}{\sqrt{10}}$ Let $u = \tan(x)$. The equation becomes a quadratic: $u^2 - 2u - 3 = 0$. Factoring gives $(u-3)(u+1) = 0$, so $u=3$ or $u=-1$. This means $\tan(x) = 3$ or $\tan(x) = -1$. Since $x$ is an acute angle (in the first quadrant), its tangent must be positive. Thus, we must have $\tan(x) = 3$. To find $\sin(x)$, we can visualize a right-angled triangle where the opposite side is 3 and the adjacent side is 1. The hypotenuse is $\sqrt{3^2 + 1^2} = \sqrt{10}$. Therefore, $\sin(x) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3}{\sqrt{10}}$.
6. For what value of $k$ is the equation $\cos^2(x) + 3\sin(x) = k$ true for $x = \frac{\pi}{6}$?
   Answer: $\frac{9}{4}$ Substitute $x = \frac{\pi}{6}$ into the expression. We know that $\sin(\frac{\pi}{6}) = \frac{1}{2}$ and $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$. The expression becomes $k = \cos^2(\frac{\pi}{6}) + 3\sin(\frac{\pi}{6}) = (\frac{\sqrt{3}}{2})^2 + 3(\frac{1}{2}) = \frac{3}{4} + \frac{3}{2}$. To add these fractions, find a common denominator: $\frac{3}{4} + \frac{6}{4} = \frac{9}{4}$.
7. Which of the following equations is a trigonometric identity?
   Answer: $\sin^2(3x) + \cos^2(3x) = 1$ When you encounter questions about trigonometric identities, you're looking for equations that are true for all values in the domain where the functions are defined, not just specific values of the variable. The key insight here is recognizing the fundamental Pythagorean identity: $\sin^2(\theta) + \cos^2(\theta) = 1$. This identity holds for any angle $\theta$, whether it's a simple variable like $x$ or a more complex expression like $3x$. In option D, we have $\sin^2(3x) + \cos^2(3x) = 1$, which is exactly this fundamental identity with $\theta = 3x$. This equation is true for all real values of $x$. Option A, $\sin(x) + \cos(x) = 1$, is only true for specific values of $x$, not all values. For example, when $x = 0$, we get $\sin(0) + \cos(0) = 0 + 1 = 1$, but when $x = \frac{\pi}{2}$, we get $\sin(\frac{\pi}{2}) + \cos(\frac{\pi}{2}) = 1 + 0 = 1$. However, for $x = \frac{\pi}{4}$, we get $\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} \neq 1$. Option B, $\tan(x) = 1$, is only true when $x = \frac{\pi}{4} + n\pi$ for integer values of $n$. Option C, $\cos(x) = \sqrt{1 - \sin^2(x)}$, is problematic because the square root only gives the positive value, but cosine can be negative. Remember: trigonometric identities must work for all valid input values. The Pythagorean identity $\sin^2(\theta) + \cos^2(\theta) = 1$ is your most reliable tool for recognizing these universal relationships.
8. The expression $(\cos(A) - \sin(A))^2 - 1$ simplifies to:
   Answer: $-2\sin(A)\cos(A)$ First, expand the squared term: $(\cos(A) - \sin(A))^2 = \cos^2(A) - 2\sin(A)\cos(A) + \sin^2(A)$. Now substitute this back into the original expression: $\cos^2(A) - 2\sin(A)\cos(A) + \sin^2(A) - 1$. Using the Pythagorean identity $\cos^2(A) + \sin^2(A) = 1$, we can simplify the expression to $1 - 2\sin(A)\cos(A) - 1$, which equals $-2\sin(A)\cos(A)$.
9. Given that $3\sin^2(\theta) + 7\cos(\theta) = 5$ for $0 < \theta < \frac{\pi}{2}$, find the value of $\cos(\theta)$.
   Answer: $\frac{1}{3}$ Use the identity $\sin^2(\theta) = 1 - \cos^2(\theta)$ to write the equation solely in terms of $\cos(\theta)$. $3(1 - \cos^2(\theta)) + 7\cos(\theta) = 5$. Expand and rearrange: $3 - 3\cos^2(\theta) + 7\cos(\theta) - 5 = 0$, which simplifies to $-3\cos^2(\theta) + 7\cos(\theta) - 2 = 0$. Multiply by -1 to get $3\cos^2(\theta) - 7\cos(\theta) + 2 = 0$. Let $c = \cos(\theta)$. The quadratic is $3c^2 - 7c + 2 = 0$. Factoring gives $(3c - 1)(c - 2) = 0$. The solutions are $c = \frac{1}{3}$ or $c = 2$. Since the range of $\cos(\theta)$ is $[-1, 1]$, the solution $\cos(\theta) = 2$ is impossible. Therefore, $\cos(\theta) = \frac{1}{3}$. This is a valid solution as it is positive, consistent with $\theta$ being in the first quadrant.
10. Given that $\sin(x) = p$ where $\frac{\pi}{2} < x < \pi$, express $\tan(x)$ in terms of $p$.
    Answer: $-\frac{p}{\sqrt{1-p^2}}$ First, find $\cos(x)$ using $\sin^2(x) + \cos^2(x) = 1$. We have $\cos^2(x) = 1 - \sin^2(x) = 1 - p^2$, so $\cos(x) = \pm\sqrt{1-p^2}$. The given domain $\frac{\pi}{2} < x < \pi$ is the second quadrant, where cosine is negative. Therefore, $\cos(x) = -\sqrt{1-p^2}$. Now, use the identity $\tan(x) = \frac{\sin(x)}{\cos(x)}$. Substituting the known values gives $\tan(x) = \frac{p}{-\sqrt{1-p^2}} = -\frac{p}{\sqrt{1-p^2}}$.
11. Which of the following expressions is equivalent to $\frac{1 - \cos(\theta)}{\sin(\theta)}$?
    Answer: $\frac{\sin(\theta)}{1 + \cos(\theta)}$ To find an equivalent expression, we can multiply the numerator and denominator by the conjugate of the numerator, which is $1 + \cos(\theta)$. This gives $\frac{1 - \cos(\theta)}{\sin(\theta)} \times \frac{1 + \cos(\theta)}{1 + \cos(\theta)} = \frac{1 - \cos^2(\theta)}{\sin(\theta)(1 + \cos(\theta))}$. Using the Pythagorean identity $\sin^2(\theta) + \cos^2(\theta) = 1$, we know that $1 - \cos^2(\theta) = \sin^2(\theta)$. The expression becomes $\frac{\sin^2(\theta)}{\sin(\theta)(1 + \cos(\theta))}$. Canceling one factor of $\sin(\theta)$ from the numerator and denominator yields $\frac{\sin(\theta)}{1 + \cos(\theta)}$.
12. Given that $\tan(\theta) = k$ for an acute angle $\theta$, which expression represents the product $\sin(\theta)\cos(\theta)$?
    Answer: $\frac{k}{k^2+1}$ When you encounter trigonometric problems involving one ratio and need to find another, the key strategy is to use the Pythagorean identity and the fundamental relationships between sine, cosine, and tangent. Given that $\tan(\theta) = k$ where $\theta$ is acute, you can construct a right triangle where the opposite side is $k$ and the adjacent side is $1$. Using the Pythagorean theorem, the hypotenuse equals $\sqrt{k^2 + 1}$. From this triangle: $\sin(\theta) = \frac{k}{\sqrt{k^2+1}}$ and $\cos(\theta) = \frac{1}{\sqrt{k^2+1}}$ Therefore: $\sin(\theta)\cos(\theta) = \frac{k}{\sqrt{k^2+1}} \cdot \frac{1}{\sqrt{k^2+1}} = \frac{k}{k^2+1}$ This confirms answer D is correct. Answer A, $\frac{1}{k^2+1}$, represents $\cos^2(\theta)$, not the product $\sin(\theta)\cos(\theta)$. This is a common error when students confuse which trigonometric expression they're calculating. Answer B, $\frac{k}{k^2-1}$, uses subtraction instead of addition in the denominator. This mistake often occurs when students misremember the Pythagorean identity or confuse it with other algebraic identities. Answer C, $\frac{k}{\sqrt{k^2+1}}$, represents $\sin(\theta)$ alone, not the product. Students might select this if they forget to multiply by the cosine term. Remember: when working with trigonometric ratios, always draw a reference triangle to visualize the relationships. This prevents algebraic errors and helps you verify your final expression makes geometric sense.
13. The expression $\cos(x) + \sin(x)\tan(x)$ is equivalent to:
    Answer: $\frac{1}{\cos(x)}$ Substitute $\tan(x) = \frac{\sin(x)}{\cos(x)}$ into the expression: $\cos(x) + \sin(x) \left( \frac{\sin(x)}{\cos(x)} \right) = \cos(x) + \frac{\sin^2(x)}{\cos(x)}$. To add these terms, find a common denominator: $\frac{\cos^2(x)}{\cos(x)} + \frac{\sin^2(x)}{\cos(x)} = \frac{\cos^2(x) + \sin^2(x)}{\cos(x)}$. Using the Pythagorean identity $\sin^2(x) + \cos^2(x) = 1$, the expression simplifies to $\frac{1}{\cos(x)}$.
14. If $\sin(x) - \cos(x) = \frac{1}{3}$, find the value of $\sin(x)\cos(x)$.
    Answer: $\frac{4}{9}$ Square both sides of the equation $\sin(x) - \cos(x) = \frac{1}{3}$: $(\sin(x) - \cos(x))^2 = (\frac{1}{3})^2$. Expand the left side: $\sin^2(x) - 2\sin(x)\cos(x) + \cos^2(x) = \frac{1}{9}$. Rearrange and use the identity $\sin^2(x) + \cos^2(x) = 1$: $(\sin^2(x) + \cos^2(x)) - 2\sin(x)\cos(x) = \frac{1}{9}$. This becomes $1 - 2\sin(x)\cos(x) = \frac{1}{9}$. Rearranging to solve for $\sin(x)\cos(x)$: $1 - \frac{1}{9} = 2\sin(x)\cos(x)$, so $\frac{8}{9} = 2\sin(x)\cos(x)$. Finally, divide by 2: $\sin(x)\cos(x) = \frac{4}{9}$.
15. The expression $\frac{1 - \sin^2(x)}{1 - \cos^2(x)}$ is equivalent to:
    Answer: $\cot^2(x)$ Using the Pythagorean identity $\sin^2(x) + \cos^2(x) = 1$, we can rewrite the numerator and the denominator. The numerator $1 - \sin^2(x)$ is equal to $\cos^2(x)$. The denominator $1 - \cos^2(x)$ is equal to $\sin^2(x)$. So the expression becomes $\frac{\cos^2(x)}{\sin^2(x)}$. This is equal to $\left(\frac{\cos(x)}{\sin(x)}\right)^2$. Since $\cot(x) = \frac{\cos(x)}{\sin(x)}$, the expression simplifies to $\cot^2(x)$.
16. Let $f(x) = (\sin(x) + \cos(x))^2$. Find the range of $f(x)$.
    Answer: $[0, 2]$ Expand the expression for $f(x)$: $f(x) = \sin^2(x) + 2\sin(x)\cos(x) + \cos^2(x)$. Using the identity $\sin^2(x) + \cos^2(x) = 1$, this simplifies to $f(x) = 1 + 2\sin(x)\cos(x)$. Using the double angle identity $\sin(2x) = 2\sin(x)\cos(x)$, we get $f(x) = 1 + \sin(2x)$. The range of the function $\sin(2x)$ is $[-1, 1]$. Therefore, the range of $f(x) = 1 + \sin(2x)$ is $[1 + (-1), 1 + 1]$, which is $[0, 2]$.
17. Simplify the expression $\frac{\sin^4(\theta) - \cos^4(\theta)}{\sin^2(\theta) - \cos^2(\theta)}$, assuming the denominator is not zero.
    Answer: 1 The numerator $\sin^4(\theta) - \cos^4(\theta)$ is a difference of two squares, which can be factored as $(\sin^2(\theta) - \cos^2(\theta))(\sin^2(\theta) + \cos^2(\theta))$. The expression becomes $\frac{(\sin^2(\theta) - \cos^2(\theta))(\sin^2(\theta) + \cos^2(\theta))}{\sin^2(\theta) - \cos^2(\theta)}$. The term $(\sin^2(\theta) - \cos^2(\theta))$ cancels out, leaving $\sin^2(\theta) + \cos^2(\theta)$. By the Pythagorean identity, this is equal to 1.
18. If $x$ is an angle in the second quadrant such that $\tan(x) = -\frac{4}{3}$, what is the value of $5(\sin(x) + \cos(x))$?
    Answer: 1 In the second quadrant, $\sin(x)$ is positive and $\cos(x)$ is negative. From $\tan(x) = -\frac{4}{3}$, we can model a right-angled triangle with an opposite side of 4 and an adjacent side of 3. The hypotenuse is $\sqrt{4^2 + 3^2} = \sqrt{25} = 5$. Using the quadrant information for the signs, we have $\sin(x) = \frac{4}{5}$ and $\cos(x) = -\frac{3}{5}$. Now substitute these values into the expression: $5(\sin(x) + \cos(x)) = 5(\frac{4}{5} + (-\frac{3}{5})) = 5(\frac{4}{5} - \frac{3}{5}) = 5(\frac{1}{5}) = 1$.
19. The expression $\frac{1}{1 - \sin(x)} + \frac{1}{1 + \sin(x)}$ is equivalent to:
    Answer: $\frac{2}{\cos^2(x)}$ To add the fractions, find a common denominator, which is $(1 - \sin(x))(1 + \sin(x))$. This product simplifies to $1 - \sin^2(x)$, which by the Pythagorean identity is equal to $\cos^2(x)$. The numerator becomes $(1 + \sin(x)) + (1 - \sin(x)) = 2$. Therefore, the expression is $\frac{2}{\cos^2(x)}$.

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