Identities with Angle Sums

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Trigonometry › Identities with Angle Sums

Questions 1 - 10

Given $\phi = \pi$ , what is $\tan\left ( \theta + \phi \right )$ ?

$\tan\left ( \theta\right )$

$\cos\left ( \theta\right )$

$\sin\left ( \theta\right )$

Explanation

We need to use the formula

$\tan\left ( \theta \pm \phi\right ) = \frac{\tan(\theta) \pm \tan(\phi)}{1 \mp \tan(\theta)\tan(\phi)}$

Substituting $\phi = \pi$ , and $\tan (\pi) = 0$ ,

$\tan\left ( \theta + \pi\right ) = \frac{\tan(\theta) + \tan(\phi)}{1 - \tan(\theta)\tan(\phi)} = \frac{\tan(\theta) + 0}{1 - \tan(\theta)\left ( 0\right )} = \tan(\theta)$

Find the exact value of $\cos 105^{\circ}$ using $105^{\circ} = 60^{\circ}+45^{\circ}$ .

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

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

$\frac{-(\sqrt{2}+\sqrt{6})}{4}$

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

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

Explanation

Our basic sum formula for cosine is:

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

Substituting the relevant angles gives us:

$\cos 105^{\circ}=\cos 60^{\circ}\cos 45^{\circ} - \sin 60^{\circ}\sin 45^{\circ}$

Now substitute in the exact values for each function, simplifying to keep radicals out of the denominator:

$\cos 105^{\circ} = \frac{1}{2}\cdot \frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{2}\cdot \frac{\sqrt {2}}{2}$

Multiply and subtract to obtain:

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

Find the exact value of the expression:

$\sin 60^{\circ}\cos 30^{\circ} + \cos 60^{\circ} \sin 30^{\circ}$

The expression is undefined.

$\frac{1}{2}$

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

Explanation

There are two ways to solve this problem. If one recognizes the identity

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

the answer is as simple as:

$\sin 60^{\circ}\cos 30^{\circ} + \cos 60^{\circ}\sin 30^{\circ} = \sin (60^{\circ}+30^{\circ}) = \sin 90^{\circ} = 1$

If one misses the identity, or wishes to be more thorough, you can simplify:

$\sin 60^{\circ}\cos 30^{\circ} + \cos 60^{\circ}\sin 30^{\circ}$

$=\bigg (\bigg(\frac{\sqrt{3}}{2}\bigg)\cdot\bigg(\frac{\sqrt{3}}{2}\bigg)\bigg)+\bigg(\bigg(\frac{1}{2}\bigg)\cdot\bigg(\frac{1}{2}\bigg)\bigg)=\frac{3}{4} + \frac{1}{4}=1$

Find the exact value of the expression:

$\cos\bigg(\frac{5\pi }{6}\bigg)\cos\bigg(\frac{2\pi}{3}\bigg)+\sin\bigg(\frac{5\pi}{6}\bigg)\sin\bigg(\frac{2\pi}{3}\bigg)$

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

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

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

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

Explanation

The formula for the cosine of the difference of two angles is

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

Substituting, we find that

$a= \frac{5\pi}{6}$

and

$b= \frac{2\pi}{3}$

Therefore, what we are really looking for is

$\cos (a-b)= \cos \bigg(\frac{5\pi}{6}-\frac{2\pi}{3}\bigg) = \cos\bigg(\frac{\pi}{6}\bigg) = \frac{\sqrt{3}}{2}$

Thus,

$\cos \bigg(\frac{5\pi}{6}\bigg)\cos\bigg(\frac{2\pi}{3}\bigg) + \sin\bigg (\frac{5\pi}{6}\bigg)\sin\bigg(\frac{2\pi}{3}\bigg) = \frac{\sqrt{3}}{2}$

Find the exact value of $\tan 75^{\circ}$ using $\tan 30^{\circ}$ and $\tan 45^{\circ}$ .

$2+\sqrt{3}$

$2-\sqrt{3}$

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

The quantity cannot be found exactly using the given information.

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

Explanation

The sum identity for tangent states that

$\tan (a+b) = \frac{\tan a + \tan b}{1-\tan a\tan b}$

Substituting known values for $\tan 30^{\circ}$ and $\tan 45^{\circ}$ , we have

$\tan (30^{\circ} + 45^{\circ}) = \frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}$

For ease, multiply all terms by $\sqrt{3}$ to get $\frac{\sqrt{3}+1}{\sqrt{3}-1}$ .

At this point, multiply both halves of the fraction by the conjugate of the denominator:

$\bigg(\frac{\sqrt{3}+1}{\sqrt{3}-1}\bigg)\cdot\bigg(\frac{\sqrt{3}+1}{\sqrt{3}+1}\bigg) = \frac{3+2\sqrt{3}+1}{3-1}$

Finally, simplify.

$\frac{3+2\sqrt{3}+1}{3-1} = \frac{4+2\sqrt{3}}{2} = 2+\sqrt{3}$

So, $\tan 75^{\circ} = 2+\sqrt{3}$ .

Suppose we have two angles, and , such that:

$sin\ A = 0.10$

$cos\ B = 0.10$

Furthermore, suppose that angle is located in the first quadrant and angle is located in the fourth.

What is the measure of:

Explanation

We can calculate some missing values using the pythagorean identities.

$sin^2\ B = 1 - cos^2B = 1 - 0.01 = 0.99$

$sin\ B = {0.99}^\frac{1}{2} = -0.995$

(Note the negative sign, because is in the fourth quadrant, where the sine of the angle is always negative).

$cos\ A = \sqrt{1 - sin^2A} = \sqrt{1 - 0.01} = \sqrt{0.99} = 0.995$

Note the positive value, since is in the first quadrant, where cosine is positive.

Now using the rules for double angles:

$cos\ 2B = 2\ cos^2 B - 1 = 2 * 0.1^2 - 1 = -0.98$

$sin\ 2B = 2\ cos\ B\ sinB = 2 \ * 0.1 \ * -0.995 = -0.199$

And then the angle subtraction formula:

$sin(A - 2B) = sin\ A \ cos \ 2B\ - cos\ A \ sin 2B$

$= (0.1 * -0.98) \ - (0.995 * -0.199)$

$= \textbf{0.10}$

Calculate $sin(105^\circ)$ .

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

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

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

$\frac{3\sqrt{6}+1}{4}$

$\frac{3\sqrt{6}+\sqrt3}{4}$

Explanation

Recall the formula for the sine of the sum of two angles:

Here, we can evaluate $sin(105^{\circ})$ by noticing that $105^{\circ}=60^{\circ}+45^{\circ}$ and applying the above formula to the sines and cosines of these two angles.