The addition formulae that are used to find trigonometry ratios of addition angles are as follows:
\(\text{sin}(A+B)= \text{sin} \ A \text{ cos} \ B + \text{cos} \ A \text{ sin} \ B\)
\(\text{sin}(A-B)= \text{sin} \ A \text{ cos} \ B - \text{cos} \ A \text{ sin} \ B\)
\(\text{cos}(A+B)= \text{cos} \ A \text{ cos} \ B - \text{sin} \ A \text{ sin} \ B\)
\(\text{cos}(A-B)= \text{cos} \ A \text{ cos} \ B + \text{sin} \ A \text{ sin} \ B\)
\(\text{tan}(A+B) = \dfrac{\text{tan }A + \text{tan }B}{1-\text{tan }A \text{ tan }B}\)
\(\text{tan}(AB) = \dfrac{\text{tan }A - \text{tan }B}{1+\text{tan }A \text{ tan }B}\)
Prove the identities for
\(\text{sin }(90^{\circ}+A) = \text{cos } A\).
Without using a calculator, find the value of \(\text{tan }15^{\circ}\).
Recall that:
\(\text{cos } 2A = \text{ cos}^2 \ A - \text{ sin}^2 \ A\)
\(\text{cos } 2A = 2\text{ cos}^2 \ A - 1\)
\(\text{cos } 2A = 1- 2\text{ sin}^2 \ A \)
Other formulae involving double angles can be derived by induction
This relation can be used to prove half-angle formulae where \(\text{sin }\dfrac{A}{2}\) , \(\text{kos }\dfrac{A}{2}\) and \(\text{tan }\dfrac{A}{2}\) are expressed in terms of \(\text{sin }A\) dan \(\text{cos }A\) as below:
\(\begin{aligned} \text{Sebelah kanan } &= \dfrac{1-\text{ cos }x}{\text{sin }x}\\\\ &= \dfrac{1-\begin{pmatrix} 1-2\text{ sin}^2 \ \dfrac{x}{2} \end{pmatrix}}{2 \text{ sin }\dfrac{x}{2}\text{ cos }\dfrac{x}{2}}\\\\ &= \dfrac{2\text{ sin}^2 \ \dfrac{x}{2} }{2 \text{ sin }\dfrac{x}{2}\text{ cos }\dfrac{x}{2}}\\\\ &=\dfrac{\text{sin }\dfrac{x}{2}}{\text{cos }\dfrac{x}{2}}\\\\ &=\text{tan }\dfrac{x}{2} \end{aligned}\)
Hence, it is proven that \(\text{tan }\dfrac{x}{2} = \dfrac{1-\text{ cos }x}{\text{sin }x}\).
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