Using Identities to Find Exact Trig Values

Solution: \(\dfrac{\sqrt{2}-\sqrt{6}}{4}\) 

Solution Method: We use sum and difference identities to find the value of a non-special angle defined by the sum or difference of two special angles. These two angles are special angles, but if I try to subtract them, the lowest common denominator would be a 12, so the difference won’t be a special angle so I can’t use my knowledge of trig values on the unit circle to calculate it. So that’s why we use a difference identity.
 

We clearly have a difference of angles and a cosine function so we’ll use the identity \[\cos{(A-B)}=\cos{A}\cos{B}+\sin{A}\sin{B}\] So we substitute \(A=\frac{5\pi}{6}\) and \(B=\frac{\pi}{4}\). \[\begin{aligned} \cos{\left(\frac{5\pi}{6}-\frac{\pi}{4}\right)}=\cos{\frac{5\pi}{6}}\cos{\frac{\pi}{4}}+\sin{\frac{5\pi}{6}}\sin{\frac{\pi}{4}} \end{aligned}\] And now we only have sine and cosine functions of special angles, which I know from the unit circle. \[\begin{aligned} \cos{\left(\frac{5\pi}{6}-\frac{\pi}{4}\right)} &= \left(-\frac{\sqrt{3}}{2}\right) \left(\frac{\sqrt{2}}{2}\right) + \left(-\frac{1}{2}\right) \left(-\frac{\sqrt{2}}{2}\right) \\ &= -\frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} \\ &= \frac{\sqrt{2}-\sqrt{6}}{4} \end{aligned}\]

Solution: \(-\dfrac{3\sqrt{3}}{10} + \dfrac{2}{5}\)
 

Solution Method: A lot of things to unpack here in this question. First thing we see is we are given the sine ratio of some angle \(\theta\). Then we see that angle \(\theta\) is between \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\). Finally, the question asks us to find \(\sin(\frac{\pi}{3}+\theta)\). So actually, we don’t need to calculate what \(\theta\) is to solve this question.

We’re asked to find sine of a sum of angles, so we’re going to use the sum identity for sine. \[\sin{(A+B)}= \sin{A}\cos{B}+\cos{A}\sin{B}\] I don’t know what \(\theta\) is but, I have the sine ratio of \(\theta\) and a domain restriction on it. Using those two facts, I can find the other trig ratios of \(\theta\), including \(\cos{\theta}\), which is what I need to use this formula.

To find \(\cos{\theta}\) given \(\sin{\theta}\), I construct a reference triangle. First, I need to know which quadrant this triangle is going in. \(\frac{\pi}{2}<\theta<\frac{3\pi}{2}\) restricts us to Q2 and Q3. Now, sine is positive in Q2, but negative in Q3, so since \(\sin{\theta}\) is equal to positive \(\frac{4}{5}\), we know we have to be in Q2.

Sine is opposite over hypotenuse so this side opposite \(\theta\) is length 4 and the hypotenuse is length 5. To find the length of this adjacent side we use pythagorean theorem. \[\begin{aligned} x^2+4^2&=5^2 \\ x^2 + 16 &= 25 \\ x^2 &= 9 \\ x &= 3 \end{aligned}\] (Of course, we also could have recognized this is a 3-4-5 triangle, and gotten there a little faster).

So the length of the adjacent side is 3, BUT we have to notice what quadrant we are in for the sign. We are in Q2, so this side is actually -3. Then, \(\cos{\theta}\) is adjacent over hypotenuse so \(\cos{\theta}=\frac{-3}{5}\)

Finally, we can rewrite \(\sin(\frac{\pi}{3}+\theta)\) using the identity as \[\sin\left(\frac{\pi}{3}+\theta\right) = \sin{\frac{\pi}{3}}\cos{\theta}+\cos{\frac{\pi}{3}}\sin{\theta}\] Then we substitute the values of \(\sin{\theta}\) and \(\cos{\theta}\), and \(\frac{\pi}{3}\) is a special angle so I know its value from the unit circle. \[\begin{aligned} \sin\left(\frac{\pi}{3}+\theta\right) &= \left(\frac{\sqrt{3}}{2}\right) \left(-\frac{3}{5}\right) + \left(\frac{1}{2}\right) \left(\frac{4}{5}\right) \\ &= -\frac{3\sqrt{3}}{10} + \frac{2}{5} \\ \end{aligned}\]

Solution: \(\dfrac{5}{2}\)
 

Solution Method: Use the Double Angle Identities to find the exact value of \(10\sin{75^\circ}\cos{75}^\circ\)

The Double Angle Identities are obtained from the sum identities in the case where the values being summed are the same. As you can see, there are multiple equivalencies for \(\cos{2x}\) due to the Pythagorean Identity.

Double Angle Identities \[\begin{aligned} \sin{2x}&=2\sin{x}\cos{x}\\ \cos{2x}&=\cos^2{x}-\sin^2{x} \\ \cos{2x}&=1-2\sin^2{x} \\ \cos{2x}&= 2\cos^2{x}-1 \\ \tan{2x}&=\frac{2\tan{x}}{1-\tan^2{x}} \end{aligned}\]

Glancing over the Double Angle Identities, I see the closest one to the form of \(10\sin{75^\circ}\cos{75}^\circ\) is \(\sin{2x}=2\sin{x}\cos{x}\). I just need to factor out a 5 to make the coefficient a 2. \[\begin{aligned} 10\sin{75^\circ}\cos{75}^\circ &= 5(2\sin{75^\circ}\cos{75}^\circ) \\ &= 5(\sin(2(75^\circ)) \\ &= 5\sin{150^\circ} \end{aligned}\] Then 150is a special angle value so I recall from the unit circle that \[\begin{aligned} 5\sin{150^\circ} = 5 \left( \frac{1}{2} \right) = \frac{5}{2} \end{aligned}\]

Solutions: \(\sin 15^\circ = \dfrac{1}{2}\sqrt{2-\sqrt{3}},\qquad \) \(\cos 15^\circ = \dfrac{1}{2}\sqrt{2+\sqrt{3}},\qquad\) \(\tan 15^\circ = 2-\sqrt{3}\)

 

Solution Method:

Half-Angle Identities

\[\begin{aligned} \sin{\frac{x}{2}} &= \pm \sqrt{\frac{1-\cos{x}}{2}} \\ \cos{\frac{x}{2}} &= \pm \sqrt{\frac{1+\cos{x}}{2}} \\ \hspace{.32in} \tan{\frac{x}{2}} &= \frac{1-\cos{x}}{\sin{x}} = \frac{\sin{x}}{1+\cos{x}} \end{aligned}\]

\(15^\circ\) is not a special angle, but it is half of \(30^\circ\), which is. So we can use the half-angle identities to find the trig values at \(15^\circ.\) Notice for \(\sin{\frac{x}{2}}\) and \(\cos{\frac{x}{2}}\), we can have a positive or negative sign. So we have to determine what quadrant our angle is in. \(15^\circ\) is in Q1, so (ASTC), all three ratios will have positive sign.