Question

Find the value of \[ \sin^2\frac{2\pi}{3}+\cos^2\frac{5\pi}{6}-\tan^2\frac{3\pi}{4} \]

• A. \(0\)
• B. \(\frac12\)
• C. \(1\)
• D. \(\frac13\)

Hint:

Convert radian angles into standard degree angles and use exact trigonometric values. Squaring removes the negative sign for sine, cosine, and tangent values.

Answer 1

The Correct Option is B

Step 1: Evaluate \(\sin^2\frac{2\pi}{3}\).
We know that \[ \frac{2\pi}{3}=120^\circ \] So, \[ \sin\frac{2\pi}{3}=\sin120^\circ=\frac{\sqrt3}{2} \] Therefore, \[ \sin^2\frac{2\pi}{3}=\left(\frac{\sqrt3}{2}\right)^2=\frac34 \]

Step 2: Evaluate \(\cos^2\frac{5\pi}{6}\).
We know that \[ \frac{5\pi}{6}=150^\circ \] So, \[ \cos\frac{5\pi}{6}=\cos150^\circ=-\frac{\sqrt3}{2} \] Therefore, \[ \cos^2\frac{5\pi}{6}=\left(-\frac{\sqrt3}{2}\right)^2=\frac34 \]

Step 3: Evaluate \(\tan^2\frac{3\pi}{4}\).
We know that \[ \frac{3\pi}{4}=135^\circ \] So, \[ \tan\frac{3\pi}{4}=\tan135^\circ=-1 \] Therefore, \[ \tan^2\frac{3\pi}{4}=(-1)^2=1 \]

Step 4: Substitute all values.
Now, \[ \sin^2\frac{2\pi}{3}+\cos^2\frac{5\pi}{6}-\tan^2\frac{3\pi}{4} \] \[ =\frac34+\frac34-1 \] \[ =\frac32-1 \] \[ =\frac12 \]

Step 5: Final conclusion.
Therefore, \[ \boxed{\frac12} \]