Asked by Ronald
Solve each equation for exact solutions over the interval [0,360)where appropriate. Round approximate solutions to the nearest tenth degree. Sin^2Theta=Cos^2Theta+1
My Work: Using double angle Identity I subtract 1 to other side therefore:
1-Sin^2theta=Cos^2theta= Cos2theta=Cos^2theta.
This is where I am stuck.For this does not conclude an answer...
My Work: Using double angle Identity I subtract 1 to other side therefore:
1-Sin^2theta=Cos^2theta= Cos2theta=Cos^2theta.
This is where I am stuck.For this does not conclude an answer...
Answers
Answered by
Steve
since sin^2θ is never more than 1, you need cos^2θ = 0
so, θ = π/2 or 3π/2
or, solving algebraically,
sin^2θ = cos^2θ + 1
cos^2θ - sin^2θ = -1
cos(2θ) = -1
θ = π/2 or 3π/2
so, θ = π/2 or 3π/2
or, solving algebraically,
sin^2θ = cos^2θ + 1
cos^2θ - sin^2θ = -1
cos(2θ) = -1
θ = π/2 or 3π/2
Answered by
Cabdimanaan
2sintheta-1=cscthet
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