解析 (sin θ cos φ)^2+(sin θ sin (φ ))^2+(cos )^2θ =(sin )^2θ (cos )^2φ+(sin )^2θ (sin )^2(φ )+(cos )^2θ =(sin )^2θ . ((cos )^2(φ )+(sin )^2(φ )). +(cos )^2θ =(sin )^2θ +(cos )^2θ =1 ...
View Solution The values of θ lying between 0 and π/2 and satisfying the equation∣∣1+sin2θsin2θsin2θcos2θ1+cos2θcos2θ4sin4θ4sin4θ1+4sin4θ∣∣=0 are View Solution Free Ncert Solutions English Medium NCERT Solutions ...
Answer to: Prove the trigonometric identity \displaystyle{ \left( \sin(x) + \cos(x) \right)^2 \sin(2x) = 1. } By signing up, you'll get...
Prove that sin2+cos2=1. Pythagorean Identity The Pythagorean identity states that the square of sine of a particular angle added to the square of cosine of same angle always results in 1. The proof of this identity is based on simple geometrical interpretation of a right triangle. Answer and...
【解析】 Consider the identity$$ \frac { \cot ^ { 2 } - \sin \theta } { \sin t \cos t } = \cot t - \tan t $$ T o verify the identity, start with left side an d transform it into cott-tant Simplify the left side of equation by dividing each term of numerator by ...
Verify the Identity cos(4u)=cos(2u)^2-sin(2u)^2 ( (cos)(4u)=((cos))^2(2u)-((sin))^2(2u)) 相关知识点: 试题来源: 解析 The provided equation is an identity but there are no steps available. ( (cos)(4u)=((cos))^2(2u)-((sin))^2(2u)) is an identity反馈 收藏 ...
sin(π2)cos(x)+cos(π2)sin(x)sin(π2)cos(x)+cos(π2)sin(x)Simplify the expression. Tap for more steps... cos(x)cos(x)Because the two sides have been shown to be equivalent, the equation is an identity. sin(π2+x)=cos(x)sin(π2+x)=cos(x) is an identitysin(π2+x)...
Answer to: Verify the identity. sin x plus or minus sin y / cos x + cos y = tan x plus or minus y / 2 By signing up, you'll get thousands of...
Answer to: Verify the identity. cot x cos x + sin x = csc x By signing up, you'll get thousands of step-by-step solutions to your homework...
( cos θ + sin θ )^2=1+ sin 2 θ 相关知识点: 试题来源: 解析 1+sin2θ=1+sin2θ ( cos θ + sin θ )^2=1+ sin 2 θcos^2θ +2sinθ cosθ +sin^2θ =1+sin2θ 1+2sinθ cosθ =1+sin2θ 1+sin2θ =1+sin2θ ...