Question

The value of $\sin^{4}\frac{\pi}{8}+\sin^{4}\frac{3\pi}{8}$ is equal to

• A. $\frac{5}{8}$
• B. $\frac{3}{4}$
• C. $\frac{3}{\sqrt{2}}$
• D. $\frac{3}{8}$
• E. $\frac{5}{4}$

Hint:

Algebra Tip: Memorize the identity $\sin^4\theta + \cos^4\theta = 1 - \frac{1}{2}\sin^2(2\theta)$. It appears constantly in advanced trigonometry problems and skips three lines of algebraic working!

Answer 1

The Correct Option is B

Concept:
Use complementary angles to convert one sine term to a cosine term, then apply the algebraic identity $a^4 + b^4 = (a^2 + b^2)^2 - 2a^2b^2$ alongside the double angle formula for sine, $\sin(2\theta) = 2\sin\theta\cos\theta$.

Step 1: Relate the angles using complementary identities.
Notice that $\frac{3\pi}{8}$ and $\frac{\pi}{8}$ add up to $\frac{4\pi}{8}$, which is $\frac{\pi}{2}$ (or $90^\circ$). Therefore, $\frac{3\pi}{8} = \frac{\pi}{2} - \frac{\pi}{8}$. This allows us to use the co-function identity $\sin\left(\frac{\pi}{2} - \theta\right) = \cos\theta$.

Step 2: Rewrite the expression.
Substitute the cosine term into the original expression: $$\sin^4\left(\frac{\pi}{8}\right) + \sin^4\left(\frac{\pi}{2} - \frac{\pi}{8}\right)$$ $$= \sin^4\left(\frac{\pi}{8}\right) + \cos^4\left(\frac{\pi}{8}\right)$$

Step 3: Apply the algebraic identity for 4th powers.
Using $a^4 + b^4 = (a^2 + b^2)^2 - 2a^2b^2$, where $a = \sin\frac{\pi}{8}$ and $b = \cos\frac{\pi}{8}$: $$= \left(\sin^2\frac{\pi}{8} + \cos^2\frac{\pi}{8}\right)^2 - 2\sin^2\left(\frac{\pi}{8}\right)\cos^2\left(\frac{\pi}{8}\right)$$ Since $\sin^2\theta + \cos^2\theta = 1$, the first term is simply $1^2 = 1$.

Step 4: Use the double angle formula.
The remaining expression is $1 - 2\sin^2\left(\frac{\pi}{8}\right)\cos^2\left(\frac{\pi}{8}\right)$. We can rewrite the second term using $\sin(2\theta) = 2\sin\theta\cos\theta$, which means $\sin^2\theta\cos^2\theta = \frac{\sin^2(2\theta)}{4}$: $$= 1 - 2 \left( \frac{\sin^2(2 \cdot \frac{\pi}{8})}{4} \right)$$ $$= 1 - \frac{1}{2} \sin^2\left(\frac{\pi}{4}\right)$$

Step 5: Substitute standard values and simplify.
We know that $\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$. Substitute this into the expression: $$= 1 - \frac{1}{2} \left( \frac{1}{\sqrt{2}} \right)^2$$ $$= 1 - \frac{1}{2} \left( \frac{1}{2} \right)$$ $$= 1 - \frac{1}{4} = \frac{3}{4}$$ Hence the correct answer is (B) $\frac{3{4}$}.