Question

Two projectiles are projected at angles ( pi / 4+theta) and pi/4+theta with the horizontal, where theta

Answer 1

solution:

from the compound angle formulae for the tangent function, we have:

tan(A+B)=tan(A)+tan(B)1−tan(A)tan(B)tan(A+B)=tan(A)+tan(B)1−tan(A)tan(B)

tan(A−B)=tan(A)−tan(B)1+tan(A)tan(B)tan(A−B)=tan(A)−tan(B)1+tan(A)tan(B)

Substituting A=π4⇒tan(A)=1A=π4⇒tan(A)=1

and B=θB=θ, we have:

X=tan(π4+θ)=1+tan(θ)1−tan(θ)X=tan(π4+θ)=1+tan(θ)1−tan(θ)

Y=tan(π4−θ)=1−tan(θ)1+tan(θ)Y=tan(π4−θ)=1−tan(θ)1+tan(θ)

Subtracting YY from XX, we have:

X−Y=1+tan(θ)1−tan(θ)−1−tan(θ)1+tan(θ)X−Y=1+tan(θ)1−tan(θ)−1−tan(θ)1+tan(θ)

We will now express the right side over a common denominator:

X−Y=1+tan(θ)1−tan(θ)×1+tan(θ)1+tan(θ)−1−tan(θ)1+tan(θ)×1−tan(θ)1−tan(θ)X−Y=1+tan(θ)1−tan(θ)×1+tan(θ)1+tan(θ)−1−tan(θ)1+tan(θ)×1−tan(θ)1−tan(θ)

=4tan(θ)1−tan2(θ)=4tan(θ)1−tan2(θ)

From the double angle formula for tan,

X−Y=2tan(2θ)

Answer 2

Step-by-step explanation:

sin(π/2+α)=cosα

sin(π/2-α)=cosα