Solving Trigonometric Equations and Inequalities

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Pre-Calculus › Solving Trigonometric Equations and Inequalities

Questions 1 - 10

Find the value of .

$\frac{\sqrt2}{2}$

$\sqrt2$

$\frac{\sqrt6+\sqrt2}{2}$

$\sqrt6+\sqrt2$

Explanation

To solve , we will need to use both the sum and difference identities for cosine.

Write the formula for these identities.

$cos(A\pm B)= cos(A)cos(B)\pm sin(A)sin(B)$

To solve for and , find two special angles whose difference and sum equals to the angle 15 and 75, respectively. The two special angles are 45 and 30.

Substitute the special angles in the formula.

$cos(45\pm 30)= cos(45)cos(30)\pm sin(45)sin(30)$

Evaluate both conditions.

$=\frac{\sqrt2}{2}\left(\frac{\sqrt3}{2}\right)+\frac{\sqrt2}{2}\left(\frac{1}{2}\right)=\frac{\sqrt6}{4}+\frac{\sqrt2}{4}$

$=\frac{\sqrt2}{2}\left(\frac{\sqrt3}{2}\right)-\frac{\sqrt2}{2}\left(\frac{1}{2}\right)=\frac{\sqrt6}{4}-\frac{\sqrt2}{4}$

Solve for .

$=\frac{\sqrt6}{4}+\frac{\sqrt2}{4}-\left[\frac{\sqrt6}{4}-\frac{\sqrt2}{4}\right]$

$=\frac{\sqrt6}{4}+\frac{\sqrt2}{4}-\frac{\sqrt6}{4}+\frac{\sqrt2}{4}$

$= \frac{2\sqrt2}{4}$

$= \frac{\sqrt2}{2}$

Evaluate

$\sin \left(\frac{5 \pi}{12}\right)$ .

$\frac{1+\sqrt3}{2 \sqrt 2 }$

$\frac{-1+\sqrt3}{2 \sqrt 2 }$

$\frac{1}{\sqrt 2 }$

$\sqrt{\frac{3}{2}}$

$\frac{-1 - \sqrt 3 }{\sqrt 2 }$

Explanation

$\frac{5 \pi }{12}$ is equivalent to $\frac{2 \pi }{12 } + \frac{3 \pi }{12 }$ or more simplified $\frac{\pi}{6 } + \frac{\pi }{4}$ .

We can use the sum identity to evaluate this sine:

$\sin \left(\frac{\pi }{6} + \frac{\pi }{4 } \right) = \sin\left (\frac{\pi }{6} \right) \cos \left(\frac { \pi }{4} \right) + \cos \left(\frac{\pi }{6}\right) \sin \left( \frac{ \pi }{4} \right)$

From the unit circle, we can determine these measures:

$\frac{1}{2} \cdot \frac{1}{\sqrt 2 } + \frac{ \sqrt 3 }{2} \cdot \frac{1}{ \sqrt 2 } = \frac{1}{2 \sqrt 2 }+ \frac{\sqrt 3 }{2 \sqrt 2 } = \frac{ 1 + \sqrt 3 }{ 2 \sqrt 2 }$

Find $\sin\bigg(\frac{\pi }{3} + \frac{\pi }{6}\bigg)$ using the sum identity.

$\frac{\sqrt{3}}{2}$

$\frac{\pi }{2}$

Explanation

Using the sum formula for sine,

$\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$

where,

$a=\frac{\pi}{3}$ , $b=\frac{\pi}{6}$

yeilds:

$\sin \bigg(\frac{\pi}{3}+\sin\frac{\pi}{6}\bigg)=\sin\bigg(\frac{\pi }{3}\bigg)\cdot\cos\bigg(\frac{\pi }{6}\bigg) +\cos \bigg(\frac{\pi}{3}\bigg)\cdot\sin\bigg(\frac{\pi }{6}\bigg)$

$=\frac{\sqrt{3}}2{}\cdot\frac{\sqrt{3}}2{}+\frac{1}{2}\cdot\frac{1}{2}$

$=\frac{3}{4}+\frac{1}{4}=1$ .

In the problem below, $\sin \alpha = \frac {1}{5}, 0 < \alpha < \frac {\pi}{2}$ and $\cos \beta = \frac {1}{4}, 0 < \beta< \frac {\pi}{2}$ .

Find

$\sin (\alpha + \beta)$ .

$\frac {1+ 6\sqrt{10}}{20}$

$\frac {1- 6\sqrt{10}}{20}$

$\frac {\sqrt{6} - \sqrt{15}}{10}$

$\frac {2\sqrt{6} - \sqrt{15}}{20}$

$\frac {2\sqrt{6} + \sqrt{15}}{20}$

Explanation

Since $\sin \alpha = \frac {1}{5}$ and $\alpha$ is in quadrant I, we can say that and and therefore:

$x = \sqrt {r^2 - y^2} = \sqrt {25-1} = \sqrt {24} = 2 \sqrt {6}$ .

So $\cos \alpha = \frac {2\sqrt{6}}{5}$ .

Since $\cos \beta = \frac {1}{4}$ and $\beta$ is in quadrant I, we can say that and and therefore:

$y = \sqrt {r^2 - x^2} = \sqrt {16-1} = \sqrt {15}$ .

So $\sin \beta = \frac {\sqrt{15}}{4}$ .

Using the sine sum formula, we see:

$\ \sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \ \* = \frac {1}{5} \cdot \frac {1}{4} + \frac {2 \sqrt {6}}{5} \cdot \frac {\sqrt {15}}{4} = \frac {1 + 6 \sqrt {10}}{20}$

Solve $\sin^2(x) + 4 = 0$ for $0^{\circ}\leq x\leq 360^{\circ}$

There is no solution.

$x = 90^{\circ}$

Explanation

By subtracting from both sides of the original equation, we get $\sin^2(x) = -4$ . We know that the square of a trigonometric identity cannot be negative, regardless of the input, so there can be no solution.

Solve $\sin^2(x) + 4 = 0$ for $0^{\circ}\leq x\leq 360^{\circ}$

There is no solution.

$x = 90^{\circ}$

Explanation

In the problem below, $\sin \alpha = \frac {1}{5}, 0 < \alpha < \frac {\pi}{2}$ and $\cos \beta = \frac {1}{4}, 0 < \beta< \frac {\pi}{2}$ .

Find

$\cos (\alpha - \beta)$ .

$\frac {2\sqrt{6} + \sqrt{15}}{20}$

$\frac {2\sqrt{6} - \sqrt{15}}{20}$

$\frac {1+ 6\sqrt{10}}{20}$

$\frac {1- 6\sqrt{10}}{20}$

$\frac {\sqrt{6} - \sqrt{15}}{10}$

Explanation

Since $\sin \alpha = \frac {1}{5}$ and $\alpha$ is in quadrant I, we can say that and and therefore:

$x = \sqrt {r^2 - y^2} = \sqrt {25-1} = \sqrt {24} = 2 \sqrt {6}$ .

So $\cos \alpha = \frac {2\sqrt{6}}{5}$ .

Since $\cos \beta = \frac {1}{4}$ and $\beta$ is in quadrant I, we can say that and and therefore:

$y = \sqrt {r^2 - x^2} = \sqrt {16-1} = \sqrt {15}$ .

So $\sin \beta = \frac {\sqrt{15}}{4}$ .

Using the cosine difference formula, we see:
$\ \cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta\ \* = \frac {2 \sqrt {6}}{5} \cdot \frac {1}{4} + \frac {1}{5} \cdot \frac {\sqrt {15}}{4} = \frac {2 \sqrt{6} + \sqrt {15}}{20}$

Given that $\tan x = 2$ and $\tan y = 3$ , find $\tan (x+y)$ .

Explanation

Jump straight to the tangent sum formula:

$\ \tan (x+y) = \frac {\tan x + \tan y}{1-\tan x \tan y}$

From here plug in the given values and simplify.

$\ \ = \frac {2+3}{1-2\cdot3} \ \= \frac {5}{-5} \ \= -1$

Calculate $\sin(75)$ .

$\frac{\sqrt{2}+\sqrt{6}}{4}$

$\frac{6+\sqrt{2}}{4}$

$\frac{1}2{}$

Explanation

Notice that $\sin(75)$ is equivalent to $\sin(30+45)$ . With this conversion, the sum formula can be applied using,

$\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$

where

, .

Therefore the result is as follows:

$\sin(a+b)=\sin(30)\cdot \cos(45)+\cos(30)\cdot\sin(45)$

$=\frac{1}{2}\cdot \frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}2{}$

$=\frac{\sqrt2}{4}+\frac{\sqrt6}{4}=\frac{\sqrt2 +\sqrt6}{4}$ .

Use trigonometric identities to solve the equation for the angle value.

$4\cos^2\theta+\cos{2\theta}+\sin^2\theta=3$

$\theta\approx39.232$

$\theta\approx78.463$

$\theta=\frac{\pi}{2}$

$\theta=\frac{\pi}{6}$

Explanation

The simplest way to solve this problem is using the double angle identity for cosine. $\cos2\theta=\cos^2\theta-\sin^2\theta$

Substituting this value into the original equation gives us:

$4\cos^2\theta+\cos^2\theta-\sin^2\theta+\sin^2\theta=3\ \implies5\cos^2\theta=3\implies\cos^2\theta=\frac{3}{5}\ \cos\theta=\frac{\sqrt15}{5}\implies\theta=\cos^{-1}\frac{\sqrt15}{5}}\approx39.232$

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