The following equation is proved: (1) cos (x + 2 / π) = - SiNx (2) sin (π - x) = SiNx

The following equation is proved: (1) cos (x + 2 / π) = - SiNx (2) sin (π - x) = SiNx

Methods: first, we can use the unit circle to determine x as the acute angle, and then judge according to the positive and negative situation of SiNx and cosx in each quadrant (detailed in this senior high school textbook). Second, we can also use the difference product to directly decompose it

(SiNx + sin θ). (SiNx sin θ) = sin (x + θ). Cos (x - θ) proof

From the sum difference product formula, we know: SiNx + sin θ = 2Sin (x + θ) / 2cos (x - θ) / 2sinx sin θ = 2Sin (x - θ) / 2cos (x + θ) / 2, so (SiNx + sin θ); (SiNx sin θ) = 2Sin (x + θ) / 2cos [(x - θ) / 2] 2Sin (x - θ) / 2cos (x + θ) / 2 = sin (x + θ). The above conclusion of sin (x - θ) is wrong

Prove by mathematical induction: cos (x / 2) × cos (x / 2 ^ 2) ×... × cos (x / 2 ^ n) = SiNx / [2 ^ n × sin (x / 2 ^ n)]
How to prove when n = K + 1?

Let n = k, then cos (x / 2) × cos (x / 2 ^ 2) ×... × cos (x / 2 ^ k) = SiNx / [2 ^ k × sin (x / 2 ^ k)]
Then when n = K + 1, cos (x / 2) × cos (x / 2 ^ 2) ×... × cos (x / 2 ^ k) × cos [x / 2 ^ (K + 1)] = SiNx × cos [x / 2 ^ (K + 1)] / [2 ^ k × sin (x / 2 ^ k)]
By substituting sin (x / 2 ^ k) = 2 × cos [x / 2 ^ (K + 1)] × sin [x / 2 ^ (K + 1)] into the formula and eliminating cos [x / 2 ^ (K + 1)]
cos(x/2)×cos(x/2^2)×...×cos(x/2^k)×cos[x/2^(k+1)]＝sinx/[2^(k+1)×sin(x/2^(k+1))]
That is, when n = K + 1, the equation holds