Question

Find the product of : (4sinθ + 3cosθ) (4sinθ – 3cosθ)

Answer 1

`\\ \sf\longmapsto (4sin\theta+2cos\theta)(4sin\theta-3cos\theta)`

`\boxed{\sf (a+b)(a-b)=a^2-b^2}`

`\\ \sf\longmapsto (4sin\theta)^2-(3cos\theta)^2`

`\\ \sf\longmapsto 16sin^2\theta-9cos^2\theta`

Answer 2

Explanation:

(4sin0 × 4sin0) +(4sin0 × -3cos0)+ (3cos0×4sin0) +(3cos0 × -3cos0)=

16 sin²0 - 12 sin0cos0 + 12 sin0cos0 -

9cos²0

= 16sin²0 - 9 cos²0