Question

Given that cos2 θ − sin2 θ =

3/4
.
What is the value of cos θ?​

Answer 1

Answer:

cos²a-sin²a=¾

cos²a-(1-cos²a)=¾

{sin²a+cos²a=1

therefore, sin²a=1-cos²a}

cos²a-1+cos²a=¾

cos²a+cos²a=¾+1

2cos²a=7/4

cos²a=⅞

cos a=√⅞=√7/√8

hence ans is d

Answer 2

cos²θ − sin²θ = 3/4 , then the value of cos θ = √7/√8

Given : cos²θ − sin²θ = 3/4

To find : The value of cos θ

Solution :

Here it is given that

$\displaystyle \sf{ {cos}^{2} \theta - {sin}^{2} \theta = \frac{3}{4} }$

Now we have

$\displaystyle \sf{ {cos}^{2} \theta - {sin}^{2} \theta = \frac{3}{4} }$

$\displaystyle \sf{ \implies {cos}^{2} \theta - (1 - {cos}^{2} \theta )= \frac{3}{4} }$

$\displaystyle \sf{ \implies {cos}^{2} \theta - 1 + {cos}^{2} \theta = \frac{3}{4} }$

$\displaystyle \sf{ \implies 2 {cos}^{2} \theta - 1 = \frac{3}{4} }$

$\displaystyle \sf{ \implies 2 {cos}^{2} \theta = \frac{3}{4} + 1 }$

$\displaystyle \sf{ \implies 2 {cos}^{2} \theta = \frac{3 + 4}{4} }$

$\displaystyle \sf{ \implies 2 {cos}^{2} \theta = \frac{7}{4} }$

$\displaystyle \sf{ \implies {cos}^{2} \theta = \frac{7}{8} }$

$\displaystyle \sf{ \implies {cos}^{} \theta = \sqrt{ \frac{7}{8} }}$

$\displaystyle \sf{ \implies {cos}^{} \theta = \frac{ \sqrt{ 7}}{ \sqrt{ 8} }}$

The value of cos θ = √7/√8

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