Pre-Calculus - Solve Trigonometric Equations and Inequalities

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:

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

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

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

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 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 .

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

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

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

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

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:

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

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.

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

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

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:

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