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Question: If \( \sin \theta -\cos \theta =0 \), then \( {{\sin }^{4}}\theta +{{\cos }^{4}}\theta = \)? A. 1 ...

If sinθcosθ=0\sin \theta -\cos \theta =0, then sin4θ+cos4θ={{\sin }^{4}}\theta +{{\cos }^{4}}\theta =?
A. 1
B. 12\dfrac{1}{2}
C. 14\dfrac{1}{4}
D. 34\dfrac{3}{4}

Explanation

Solution

It will be very complicated to expand (a+b)4{{(a+b)}^{4}}.
Find the value of θ\theta which satisfies the given equation and then substitute it in the given expression to evaluate it.
Recall that sin45=12\sin 45{}^\circ =\dfrac{1}{\sqrt{2}} and cos45=12\cos 45{}^\circ =\dfrac{1}{\sqrt{2}}.

Complete step by step answer:
We are given the equation:
sinθcosθ=0\sin \theta -\cos \theta =0
sinθ=cosθ\sin \theta =\cos \theta
Dividing both sides by cosθ\cos \theta, we get:
sinθcosθ=1\dfrac{\sin \theta }{\cos \theta }=1
We know that sinθcosθ=tanθ\dfrac{\sin \theta }{\cos \theta }=\tan \theta, therefore, we get:
tanθ=1\tan \theta =1
We also know that tan45=1\tan 45{}^\circ =1, therefore, θ=45\theta =45{}^\circ.
Now, substituting sin45=12\sin 45{}^\circ =\dfrac{1}{\sqrt{2}} and cos45=12\cos 45{}^\circ =\dfrac{1}{\sqrt{2}} in the given expression, we get:
sin4θ+cos4θ{{\sin }^{4}}\theta +{{\cos }^{4}}\theta
= (12)4+(12)4{{\left( \dfrac{1}{\sqrt{2}} \right)}^{4}}+{{\left( \dfrac{1}{\sqrt{2}} \right)}^{4}}
= 14+14\dfrac{1}{4}+\dfrac{1}{4}
= 24\dfrac{2}{4}
= 12\dfrac{1}{2}

The correct answer is B. 12\dfrac{1}{2}.

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

Values of Trigonometric Ratios for Common Angles:

| 0°| 30°| 45°| 60°| 90°
---|---|---|---|---|---
sin| 0| 12\dfrac{1}{2} | 12\dfrac{1}{\sqrt{2}} | 32\dfrac{\sqrt{3}}{2} | 1
cos| 1| 32\dfrac{\sqrt{3}}{2} | 12\dfrac{1}{\sqrt{2}} | 12\dfrac{1}{2} | 0
tan| 0| 13\dfrac{1}{\sqrt{3}} | 1| 3\sqrt{3} | \infty
csc| \infty | 2| 2\sqrt{2} | 23\dfrac{2}{\sqrt{3}} | 1
sec| 1| 23\dfrac{2}{\sqrt{3}} | 2\sqrt{2} | 2| \infty
cot| \infty | 3\sqrt{3} | 1| 13\dfrac{1}{\sqrt{3}} | 0