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Question

Question: Find the value of \[{{\sin }^{4}}A+{{\cos }^{4}}A\], If \[sinA-cosA=0\]....

Find the value of sin4A+cos4A{{\sin }^{4}}A+{{\cos }^{4}}A, If sinAcosA=0sinA-cosA=0.

Explanation

Solution

HINT: - Before finding the solution, we must know some basic relations between the trigonometric function.
The most important relation that would be used in solving this question is as followstanx=sinxcosx\tan x=\dfrac{\sin x}{\cos x}
This above mentioned relation is the most important formula that would be used in this question.
Also, on looking at the equation that is given in the question, we can conclude that both the sin and the cos values of angle A is of the same sign, hence, as this is possible either in first or the third quadrant, therefore, we can say that angle A is either in first quadrant or in the third quadrant.

Complete step-by-step answer:
As mentioned in the question, we have to find the value of sin4A+cos4A{{\sin }^{4}}A+{{\cos }^{4}}A .
Now, as mentioned in the hint, we can solve the equation given as follows

& sinA-cosA=0 \\\ & \sin A=\cos A \\\ & \tan A=1 \\\ \end{aligned}$$ Hence, from the above equation, we can conclude that the value of angle A is either $${{45}^{\circ }}$$ or $${{225}^{\circ }}$$ . So, if the angle is $${{45}^{\circ }}$$ , then we can write the expression that is asked as follows $$\begin{aligned} & ={{\sin }^{4}}A+{{\cos }^{4}}A \\\ & ={{\sin }^{4}}{{45}^{\circ }}+{{\cos }^{4}}{{45}^{\circ }} \\\ & ={{\left( \dfrac{1}{\sqrt{2}} \right)}^{4}}+{{\left( \dfrac{1}{\sqrt{2}} \right)}^{4}}=\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2} \\\ \end{aligned}$$ Hence, the value of the expression is $$\dfrac{1}{2}$$ . Now, if the angle is $${{225}^{\circ }}$$ , then we can write the expression as follows $$\begin{aligned} & ={{\sin }^{4}}A+{{\cos }^{4}}A \\\ & ={{\sin }^{4}}{{225}^{\circ }}+{{\cos }^{4}}{{225}^{\circ }} \\\ & ={{\left( \dfrac{-1}{\sqrt{2}} \right)}^{4}}+{{\left( \dfrac{-1}{\sqrt{2}} \right)}^{4}}=\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2} \\\ \end{aligned}$$ Hence, the value of the expression is $$\dfrac{1}{2}$$ . NOTE: - The students can make an error if they don’t know about the relations between the trigonometric functions that are given in the hint as without knowing them: one could never get to the correct answer. Also, the students should know beforehand that in which quadrant, the value of which trigonometric function is positive and in which quadrant, it is negative. Also, in such questions, calculation mistakes are very common, so, to get to the right answer, one should try to be extra careful while solving these questions.