Prove (3cosθ-4sinθ)^2+(4cosθ+3sinθ)^2=25
square both terms...
9cos^2+4sin^2-24cossin+16cos^2+24cossin+9sin^2=
9(cos^2+sin^2)+16(cos^2+sin^2)=
and you can finish it.
To prove the equation (3cosθ - 4sinθ)^2 + (4cosθ + 3sinθ)^2 = 25, we will use the trigonometric identity known as the Pythagorean identity. The Pythagorean identity for cosine and sine states that:
cos^2θ + sin^2θ = 1
Let's expand the left side of the equation given:
(3cosθ - 4sinθ)^2 + (4cosθ + 3sinθ)^2
= (3cosθ - 4sinθ)(3cosθ - 4sinθ) + (4cosθ + 3sinθ)(4cosθ + 3sinθ)
Expanding and simplifying further:
= (9cos^2θ - 24cosθsinθ + 16sin^2θ) + (16cos^2θ + 24cosθsinθ + 9sin^2θ)
= 9cos^2θ - 24cosθsinθ + 16sin^2θ + 16cos^2θ + 24cosθsinθ + 9sin^2θ
= (9cos^2θ + 16cos^2θ) + (16sin^2θ + 9sin^2θ)
= 25cos^2θ + 25sin^2θ
= 25(cos^2θ + sin^2θ)
= 25 * 1 [Using the Pythagorean identity: cos^2θ + sin^2θ = 1]
= 25
Therefore, we have successfully shown that (3cosθ - 4sinθ)^2 + (4cosθ + 3sinθ)^2 equals 25, proving the given equation.