Question

d. cos^80 - sin^80 = 1/4cos20 (3 + cos 40)​

Answer 1

Step-by-step explanation:

We have,

$\tt{cos^{8} ( \theta) - sin^{8} ( \theta)}$

$\sf{ = \{cos^{4} ( \theta) - sin^{4} ( \theta) \} \{ cos^{4} ( \theta) + sin^{4} ( \theta) \}}$

$\sf{ = cos( 2\theta) \cdot [ \{ cos^{2} ( \theta) + sin^{2} ( \theta) \}^{2} - 2 sin^{2} ( \theta) cos^{2} ( \theta) } ]$

$\sf{ = cos( 2\theta) \cdot \bigg[ (1 )^{2} - \dfrac{1}{2} \cdot4 sin^{2} ( \theta) cos^{2} ( \theta) } \bigg]$

$\sf{ = cos( 2\theta) \cdot \bigg[ 1 - \dfrac{1}{2} \cdot sin^{2} ( 2\theta) } \bigg]$

$\sf{ = \dfrac{1}{4} cos( 2\theta) \cdot \bigg[ 4 - 2 sin^{2} ( 2\theta) } \bigg]$

$\sf{ = \dfrac{1}{4} cos( 2\theta) \cdot \bigg[ 4 - (1 - cos(4 \theta)) } \bigg]$

$\sf{ = \dfrac{1}{4} cos( 2\theta) \cdot \bigg[ 4 - 1 + cos(4 \theta) } \bigg]$

$\sf{ = \dfrac{1}{4} cos( 2\theta) \cdot \{ 3 + cos(4 \theta) } \}$