Test Index

AP DSC School Assistant Math Model Test 3

Section: Subject Specific

Question : 110 of 160

Marks: +1, -0

\sin ^6 \frac{\pi}{16}+\sin ^6 \frac{3 \pi}{16}+\sin ^6 \frac{5 \pi}{16}+\sin ^6 \frac{7 \pi}{16}

Solution:

Concept:

Use complementary angle identities and the algebraic identity for

a^6+b^6

to simplify the sum of sine powers.

Explanation:

Given:

\sin^6\frac{\pi}{16}+\sin^6\frac{3\pi}{16}+\sin^6\frac{5\pi}{16}+\sin^6\frac{7\pi}{16}

Note that

\sin\frac{7\pi}{16}=\cos\frac{\pi}{16}

and

\sin\frac{5\pi}{16}=\cos\frac{3\pi}{16}

. So the sum becomes

(\sin^6\frac{\pi}{16}+\cos^6\frac{\pi}{16})+(\sin^6\frac{3\pi}{16}+\cos^6\frac{3\pi}{16})

For any angle

\theta

\sin^6\theta+\cos^6\theta = (\sin^2\theta+\cos^2\theta)^3-3\sin^2\theta\cos^2\theta(\sin^2\theta+\cos^2\theta)=1-3\sin^2\theta\cos^2\theta

Thus the expression is

[1-3\sin^2\frac{\pi}{16}\cos^2\frac{\pi}{16}]+[1-3\sin^2\frac{3\pi}{16}\cos^2\frac{3\pi}{16}] = 2-3\!\left(\sin^2\frac{\pi}{16}\cos^2\frac{\pi}{16}+\sin^2\frac{3\pi}{16}\cos^2\frac{3\pi}{16}\right)

Rewrite using double-angle:

2-\frac{3}{4}\!\left[(2\sin\frac{\pi}{16}\cos\frac{\pi}{16})^2+(2\sin\frac{3\pi}{16}\cos\frac{3\pi}{16})^2\right] = 2-\frac{3}{4}\!\left(\sin^2\frac{\pi}{8}+\sin^2\frac{3\pi}{8}\right)

Now

\sin\frac{3\pi}{8}=\cos\frac{\pi}{8}

, so

\sin^2\frac{3\pi}{8}=\cos^2\frac{\pi}{8}

. Hence

\sin^2\frac{\pi}{8}+\sin^2\frac{3\pi}{8}=\sin^2\frac{\pi}{8}+\cos^2\frac{\pi}{8}=1

Therefore the value is

2-\frac{3}{4}(1)=\frac{5}{4}

Answer:

The exact value is

\frac{5}{4}

, which corresponds to Option C.

Go to Question:

Prev Question

Next Question