Question

If $ \sin \theta -\cos \theta =0 $, then $ {{\sin }^{4}}\theta +{{\cos }^{4}}\theta = $?
A. 1
B. $ \dfrac{1}{2} $
C. $ \dfrac{1}{4} $
D. $ \dfrac{3}{4} $

Answer 1

Hint: It will be very complicated to expand $ {{(a+b)}^{4}} $.
Find the value of $ \theta $ which satisfies the given equation and then substitute it in the given expression to evaluate it.
Recall that $ \sin 45{}^\circ =\dfrac{1}{\sqrt{2}} $ and $ \cos 45{}^\circ =\dfrac{1}{\sqrt{2}} $.

Complete step by step answer:
We are given the equation:
 $ \sin \theta -\cos \theta =0 $
⇒ $ \sin \theta =\cos \theta $
Dividing both sides by $ \cos \theta $, we get:
⇒ $ \dfrac{\sin \theta }{\cos \theta }=1 $
We know that $ \dfrac{\sin \theta }{\cos \theta }=\tan \theta $, therefore, we get:
⇒ $ \tan \theta =1 $
We also know that $ \tan 45{}^\circ =1 $, therefore, $ \theta =45{}^\circ $.
Now, substituting $ \sin 45{}^\circ =\dfrac{1}{\sqrt{2}} $ and $ \cos 45{}^\circ =\dfrac{1}{\sqrt{2}} $ in the given expression, we get:
 $ {{\sin }^{4}}\theta +{{\cos }^{4}}\theta $
= $ {{\left( \dfrac{1}{\sqrt{2}} \right)}^{4}}+{{\left( \dfrac{1}{\sqrt{2}} \right)}^{4}} $
= $ \dfrac{1}{4}+\dfrac{1}{4} $
= $ \dfrac{2}{4} $
= $ \dfrac{1}{2} $

The correct answer is B. $ \dfrac{1}{2} $.

Note: The value of $ \sin \theta $ and $ \cos \theta $ lies between -1 and 1.
 $ \sin (-\theta )=-\sin \theta $ and $ \cos (-\theta )=\cos \theta $.

Values of Trigonometric Ratios for Common Angles:

	0°	30°	45°	60°	90°
sin	0	$ \dfrac{1}{2} $	$ \dfrac{1}{\sqrt{2}} $	$ \dfrac{\sqrt{3}}{2} $	1
cos	1	$ \dfrac{\sqrt{3}}{2} $	$ \dfrac{1}{\sqrt{2}} $	$ \dfrac{1}{2} $	0
tan	0	$ \dfrac{1}{\sqrt{3}} $	1	$ \sqrt{3} $	$ \infty $
csc	$ \infty $	2	$ \sqrt{2} $	$ \dfrac{2}{\sqrt{3}} $	1
sec	1	$ \dfrac{2}{\sqrt{3}} $	$ \sqrt{2} $	2	$ \infty $
cot	$ \infty $	$ \sqrt{3} $	1	$ \dfrac{1}{\sqrt{3}} $	0