Question

(cos⁶theta +sin⁶theta )+3 sin²thetacos²theta​

Answer 1

Answer:

1

Step-by-step explanation:

To Find :-

Value of :-

$(cos^6\theta+sin^6\theta)+3sin^2\theta\times cos^2\theta$

How To Do :-

By using the exponential laws we need to change some terms and we need to simplify them by using a algebraic formula and then by simplifying it we need apply the value of a trigonometric identity and we need to solve it.

Formula Required :-

1) (a + b)³ = a³ + b³ + 3ab(a + b)

→ a³ + b³ = (a + b)³ - 3ab(a + b)

2) (aˣ)ⁿ = aˣⁿ

3)  sin²θ + cos²θ = 1

Solution :-

$(cos^6\theta+sin^6\theta)+3sin^2\theta\times cos^2\theta$

We can write :-

cos⁶θ = (cos²θ)³

sin⁶θ = (sin²θ)³

[∴  (aˣ)ⁿ = aˣⁿ]

Substituting the values :-

$=\left((cos^2\theta)^3+(sin^2\theta)^3)\right+3sin^2\theta\times cos^2\theta$

$= [(cos^2\theta+sin^2\theta)^3-3cos^2\theta\times sin^2\theta(cos^2\theta+sin^2\theta)]+3sin^2\theta cos^2\theta$

[ ∴ a³ + b³ = (a + b)³ - 3ab(a + b) ]

Note :- Consider , 'a' as cos²θ

'b' as sin²θ

$=[(1)^3-3cos^2\theta sin^2\theta(1)]+3sin^2\theta cos^2\theta$

[ ∴ sin²θ + cos²θ = 1 ]

= [1 - 3cos²θsin²θ] + 3sin²θcos²θ

= 1 - 3sin²θcos²θ + 3sin²θcos²θ

= 1

$\therefore (cos^6\theta+sin^6\theta)+3sin^2\theta\times cos^2\theta=1$