Question

Sin cube a+cos cube a/sina+cosa=1-1/2sin2a

Answer 1

Answer to Question #88848 – Math – Trigonometry

Question

(Sin cube a+cos cube a)/(sina+cosa)=1-1/2sin2a

Solution

As we know $(x^{3} + y^{3}) = (x + y)(x^{2} + y^{2} - xy)$

Let us assume: $x = \text{Sina And } y = \text{Cosa}$

So, putting the values of $(x,y)$ into this general expression, we get:

$$(Sin^{3}a + Cos^{3}a) = (Sina + Cosa)(Sin^{2}a + Cos^{2}a - SinaCosa)$$

So, the given expression can be simplified as:

$$\begin{array}{l} \frac{(Sin^{3}a + Cos^{3}a)}{(Sina + Cosa)} \\ = \frac{(Sina + Cosa)(Sin^{2}a + Cos^{2}a - SinaCosa)}{(Sina + Cosa)} \\ = (Sin^{2}a + Cos^{2}a - SinaCosa) \\ = (1 - SinaCosa) \quad [\text{As we know } (Sin^{2}a + Cos^{2}a) = 1] \\ = 1 - \frac{2SinaCosa}{2} \\ = 1 - \frac{Sin2a}{2} \end{array}$$

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