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Find the angle θ (if it exists) in the interval [0°, 90°) for which sin θ = cos θ
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sin θ = cos θ sin θ = sin(90-θ), so θ = 90-θ, solve for θ ?
it doesnt say. it just gives me a. θ = 30° b.θ = 45° c. No such angle exists. d. θ = 60°
yeah, it's just algebra principle... btw, i have used identity cos(90-x)=sinx
so what would be the answer
there are other ways as well --> sin/cos = tan so when is tan = 1 --> use unit circle, find angle where sin =cos
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θ = 90-θ, move -θ to left side θ+θ=90 2θ=90 divided by 2 to both sides gives θ=90/2=...
oh ok thank you!
apparently they don't want to have to do any work....unfortunately most people on this site will not just give answer without explanation ;)
agreed^ ;)
lol k...
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