Question

cos square (thita +45)+cos square (thita-45)=1​

Answer 1

Question

$\rm \: \to \: \cos {}^{2}( \theta + 45 \degree) + \cos {}^{2} ( \theta - 45 \degree) = 1$

Solution:-

Given:-

$\rm \: \to \: \cos {}^{2}( \theta + 45 \degree) + \cos {}^{2} ( \theta - 45 \degree) = 1$

Using the formula

$\rm \: \to \cos(A + B) = \cos(A) . \cos(B) - \sin(A) \sin(B)$

$\rm \: \to \cos(A - B) = \cos(A) . \cos(B) + \sin(A) \sin(B)$

We get

$\rm \: \cos {}^{2} \theta. \cos {}^{2} 45 \degree - \sin {}^{2} \theta \sin45 \degree + \cos {}^{2} \theta. \cos {}^{2} 45 \degree + \sin {}^{2} \theta \sin45 \degree = 1$

$\rm \: \cos {}^{2} \theta. \cos {}^{2} 45 \degree+ \cos {}^{2} \theta. \cos {}^{2} 45 \degree = 1$

$2 \cos {}^{2} \theta. \cos {}^{2} 45 \degree = 1$

Where

$\cos( 45 \degree) = \dfrac{1}{ \sqrt{2} }$

We get

$2 \cos {}^{2} \theta \times ( \dfrac{1}{ \sqrt{2} } ) {}^{2} = 1$

$2 \cos {}^{2} \theta \times \dfrac{1}{2} = 1$

$\cos {}^{2} \theta = 1$

$\cos \theta = 1$

$\theta = \cos {}^{ - 1} 1$

So

$\boxed{ \theta = 0 \degree}$