# Trignometry

5sin thita +12cos theta=13 then find the value of 5cos thita -12sin theta

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1. First, the name of θ is "theta" not "thita"

(5sinθ + 12cosθ)^2 = 169
25sin^2θ + 120sinθcosθ + 144cos^2θ = 169

(5cosθ - 12sinθ)^2 = 25cos^2θ - 120sinθcosθ + 144 sin^2θ
= 25cos^2θ - (169-25sin^2θ-144cos^2θ) + 144sin^2θ
= 25(cos^2θ+sin^2θ) + 144(sin^2θ+cos^2θ) - 169
= 25+144-169
= 0

or,
5sinθ + 12cosθ
= 13(5/13 sinθ + 12/13 cosθ)
now let x be the angle such that sinx = 12/13 and cosx = 5/13
= 13sin(θ+x)
So, we have
13sin(x+θ) = 13
sin(x+θ) = 1
Now, we also have
5cosθ - 12sinθ = 13(5/13 cosθ - 12/13 sinθ)
= 13cos(x+θ)
since sin(x+θ) = 1, cos(x+θ) = 0 as above

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