Fundamental Trigonometric Identities

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Pre-Calculus › Fundamental Trigonometric Identities

Questions 1 - 10

Find the exact value of each expression below without the aid of a calculator.

$sin165^{\circ}$

$\pm\sqrt\frac{\left(1-\frac{\sqrt(3)}{2}\right)}{2}$

$\pm\sqrt\frac{\left(1-\frac{1}{2}\right)}{2}$

$\pm\sqrt\frac{\left(1+\frac{\sqrt(3)}{2}\right)}{2}$

$\pm\sqrt\frac{(1+\frac{1}{2})}{2}$

$\pm1$

Explanation

In order to find the exact value of $\sin165^\circ$ we can use the half angle formula for sin, which is

$sin\left(\frac{\alpha}{2}\right) = \pm\sqrt\frac{1-cos\alpha}{2}$ .

This way we can plug in a value for alpha for which we know the exact value. $165^\circ$ is equal to $330^\circ$ divided by two, and so we can plug in $330^\circ$ for the alpha above.

The cosine of $330^\circ$ is $\frac{\sqrt3}{2}$ .

Therefore our final answer becomes,

$\pm\sqrt\frac{\left(1-\frac{\sqrt(3)}{2}\right)}{2}$ .

Which of the following is equivalent to the expression:

$\frac{1-sin^2(x)}{csc^2(x)}$

Explanation

Which of the following is equivalent to the following expression?

$\frac{1-sin^2(x)}{csc^2(x)}$

Recall our Pythagorean trig identity:

It can be rearranged to look just like our numerator:

So go ahead and change our original expression to:

$\frac{cos^2(x)}{csc^2(x)}$

Then recall the definition of cosecant:

$csc(x)=\frac{1}{sin(x)}$

So our original expression can be rewritten as:

$cos^2(x)\div\frac{1}{sin^2(x)}=cos^2(x)sin^2(x)$

So our answer is:

Find the exact value of each expression below without the aid of a calculator.

$sin165^{\circ}$

$\pm\sqrt\frac{\left(1-\frac{\sqrt(3)}{2}\right)}{2}$

$\pm\sqrt\frac{\left(1-\frac{1}{2}\right)}{2}$

$\pm\sqrt\frac{\left(1+\frac{\sqrt(3)}{2}\right)}{2}$

$\pm\sqrt\frac{(1+\frac{1}{2})}{2}$

$\pm1$

Explanation

In order to find the exact value of $\sin165^\circ$ we can use the half angle formula for sin, which is

$sin\left(\frac{\alpha}{2}\right) = \pm\sqrt\frac{1-cos\alpha}{2}$ .

This way we can plug in a value for alpha for which we know the exact value. $165^\circ$ is equal to $330^\circ$ divided by two, and so we can plug in $330^\circ$ for the alpha above.

The cosine of $330^\circ$ is $\frac{\sqrt3}{2}$ .

Therefore our final answer becomes,

$\pm\sqrt\frac{\left(1-\frac{\sqrt(3)}{2}\right)}{2}$ .

Which of the following is equivalent to the expression:

$\frac{1-sin^2(x)}{csc^2(x)}$

Explanation

Which of the following is equivalent to the following expression?

$\frac{1-sin^2(x)}{csc^2(x)}$

Recall our Pythagorean trig identity:

It can be rearranged to look just like our numerator:

So go ahead and change our original expression to:

$\frac{cos^2(x)}{csc^2(x)}$

Then recall the definition of cosecant:

$csc(x)=\frac{1}{sin(x)}$

So our original expression can be rewritten as:

$cos^2(x)\div\frac{1}{sin^2(x)}=cos^2(x)sin^2(x)$

So our answer is:

Simplify $3sin^2\theta-3$ .

$-3cos^2\theta$

$-3tan\theta$

$3-3cos^2\theta$

$3cos^2\theta$

Explanation

Write the Pythagorean Identity.

$sin^2\theta+cos^2\theta=1$

Reorganize the left side of this equation so that it matches the form: $3sin^2\theta-3$

Subtract cosine squared theta on both sides.

$sin^2\theta-1=-cos^2\theta$

Multiply both sides by 3.

$3(sin^2\theta-1=-cos^2\theta)$

$3sin^2\theta-3=-3cos^2\theta$

Simplify $3sin^2\theta-3$ .

$-3cos^2\theta$

$-3tan\theta$

$3-3cos^2\theta$

$3cos^2\theta$

Explanation

Write the Pythagorean Identity.

$sin^2\theta+cos^2\theta=1$

Reorganize the left side of this equation so that it matches the form: $3sin^2\theta-3$

Subtract cosine squared theta on both sides.

$sin^2\theta-1=-cos^2\theta$

Multiply both sides by 3.

$3(sin^2\theta-1=-cos^2\theta)$

$3sin^2\theta-3=-3cos^2\theta$

Simplify:

$\frac{2}{csc(2t)}$

$4cos(\theta)sin(\theta)$

$\frac{cos(\theta)sin(\theta)}{4}$

$2sin(\theta)$

$2cos(\theta)-2sin(\theta)$

$4sin(\theta)$

Explanation

Write the reciprocal identity for cosecant.

$csc(\theta)= \frac{1}{sin(\theta)}$

Rewrite the expression and use the double angle identities for sine to simplify.

$\\frac{2}{csc(2t)}\ \= \frac{2}{\frac{1}{sin(2\theta)}}\ \= 2sin(2\theta) \ \= 2(2sin(\theta)cos(\theta))$

$=4cos(\theta)sin(\theta)$

Simplify:

$\frac{2}{csc(2t)}$

$4cos(\theta)sin(\theta)$

$\frac{cos(\theta)sin(\theta)}{4}$

$2sin(\theta)$

$2cos(\theta)-2sin(\theta)$

$4sin(\theta)$

Explanation

Write the reciprocal identity for cosecant.

$csc(\theta)= \frac{1}{sin(\theta)}$

Rewrite the expression and use the double angle identities for sine to simplify.

$\\frac{2}{csc(2t)}\ \= \frac{2}{\frac{1}{sin(2\theta)}}\ \= 2sin(2\theta) \ \= 2(2sin(\theta)cos(\theta))$

$=4cos(\theta)sin(\theta)$

Simplify:

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

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

$-\frac{1}{2}$

$\frac{\sqrt3-1}{3}$

Explanation

Write the even and odd identities for sine and cosine.

Rewrite the expression and evaluate.

$-sin(30)+cos(30) =-\frac{1}{2}+\frac{\sqrt3}{2}= \frac{\sqrt3-1}{2}$

Simplify:

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

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

$-\frac{1}{2}$

$\frac{\sqrt3-1}{3}$

Explanation

Write the even and odd identities for sine and cosine.

Rewrite the expression and evaluate.

$-sin(30)+cos(30) =-\frac{1}{2}+\frac{\sqrt3}{2}= \frac{\sqrt3-1}{2}$

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