Using the Pythagorean identitysin2(θ)+cos2(θ)=1\sin^2(\theta) + \cos^2(\theta) =1sin2(θ)+cos2(θ)=1, we can substitutecos2(θ)\cos^2(\theta)cos2(θ)with1−sin2(θ)1-\sin^2(\theta)1−sin2(θ), obtaining ...
sin 3θ = 3 sinθ - 4 sin3θ cos 2θ = cos2θ - sin2 θ cos 2θ = 2cos2θ - 1 cos 2θ = 1- 2sin2 θ cos 3θ = 4 cos3θ - 3cosθ sin (θ/2) = ±√((1- cosθ)/2) cos (θ/2) = ±√((1+ cosθ)/2) sin θ = 2tan (θ/2) /(1 + tan2 (θ/2...
remember the identity cos(2x)=cos2(x)−sin2(x)=2cos2(x)−1. You want to solve 2cos2(x)−1+cos(x)=0. Let t=cos(x). The question 2t2+t−1=0 and is easier to solve. Can you take ... How do you simplify 2cos(10x)+4cos(5x) ? https://socratic.org/questions/ho...
(sin ^2(-θ )-cos ^2(-θ ))(sin (-θ )-cos (-θ ))=([sin (-θ )]^2-[cos (-θ )]^2)(sin (-θ )-cos (-θ ))=((-sin θ )^2-(cos θ )^2)(-sin θ -cos θ )=((sin θ -cos θ )(sin θ +cos θ ))(-(sin θ +cos θ ))=cos θ -sin θ ...
Answer to: Verify the identity. \\ \sin x+\cos x\cot x=\csc x By signing up, you'll get thousands of step-by-step solutions to your homework...
Answer to: Verify the trigonometric identity. cos(x + y) cos(x - y) = cos^2 x - sin^2 y By signing up, you'll get thousands of step-by-step...
题目Verify the Identity cot(t)-sin(2t)=cot(t)cos(2t)( (cot)(t)-(sin)(2t)=(cot)(t)(cos)(2t)) 相关知识点: 试题来源: 解析 The provided equation is an identity but there are no steps available.( (cot)(t)-(sin)(2t)=(cot)(t)(cos)(2t)) is an identity反馈 收藏 ...
Establish the identity cos + sin 0 = cos 0 - sino -1- coto 1 + tan 0 Write the left side in terms of sine and cosine. cos sin 0 1 + Write each term from the previous step as one fraction. cos 20 sin 0 - cos 0 (List the ...
1-sin(x) 1−sin(x)1-sin(x) Rewrite1−sin(x)1-sin(x)as1−1csc(x)1-1csc(x). 1−1csc(x)1-1csc(x) Because the two sides have been shown to beequivalent, theequationis anidentity. cos2(x)1+sin(x)=1−1csc(x)cos2(x)1+sin(x)=1-1csc(x)is anidentity ...
解析 (sin ^2(-θ )-cos ^2(-θ ))(sin (-θ )-cos (-θ ))=([sin (-θ )]^2-[cos (-θ )]^2)(sin (-θ )-cos (-θ ))=((-sin θ )^2-(cos θ )^2)(-sin θ -cos θ )=((sin θ -cos θ )(sin θ +cos θ ))(-(sin θ +cos θ ))=cos θ -sin ...