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受欢迎的三角函数 >

证明 4cos^2(B)-4sin^2(B)=4-8sin^2(B)

解答

证明

4 cos^{2} (B) - 4 sin^{2} (B) = 4 - 8 sin^{2} (B)

解答

真

求解步骤

4 cos^{2} (B) - 4 sin^{2} (B) = 4 - 8 sin^{2} (B)

调整左侧

4 cos^{2} (B) - 4 sin^{2} (B)

使用三角恒等式改写

4 cos^{2} (B) - 4 sin^{2} (B)

使用毕达哥拉斯恒等式:

cos^{2} (x) + sin^{2} (x) = 1

cos^{2} (x) = 1 - sin^{2} (x)

= 4 (1 - sin^{2} (B)) - 4 sin^{2} (B)

化简

4 (1 - sin^{2} (B)) - 4 sin^{2} (B) : 4 - 8 sin^{2} (B)

4 (1 - sin^{2} (B)) - 4 sin^{2} (B)

乘开

4 (1 - sin^{2} (B)) : 4 - 4 sin^{2} (B)

4 (1 - sin^{2} (B))

使用分配律:

a (b - c) = ab - a c

a = 4, b = 1, c = sin^{2} (B)

= 4 \cdot 1 - 4 sin^{2} (B)

数字相乘:

4 \cdot 1 = 4

= 4 - 4 sin^{2} (B)

= 4 - 4 sin^{2} (B) - 4 sin^{2} (B)

同类项相加:

- 4 sin^{2} (B) - 4 sin^{2} (B) = - 8 sin^{2} (B)

= 4 - 8 sin^{2} (B)

= 4 - 8 sin^{2} (B)

= 4 - 8 sin^{2} (B)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

流行的例子

证明 cos(4A)=8cos^4(A)-8cos^2(A)+1

p ro v e cos (4 A) = 8 cos^{4} (A) - 8 cos^{2} (A) + 1

证明 cos(4B)=1-8sin^2(B)+8sin^4(B)

p ro v e cos (4 B) = 1 - 8 sin^{2} (B) + 8 sin^{4} (B)

证明 sin(pi/2-u)=cos(u)

p ro v e sin (\frac{π}{2} - u) = cos (u)

证明 arccos(x)= 1/(cos(x))

p ro v e arccos (x) = \frac{1}{cos ( x )}

证明 (csc^2(x)-1)/(cos(x))=cot(x)csc(x)

p ro v e \frac{csc ^{2} ( x ) - 1}{cos ( x )} = cot (x) csc (x)