How do i part b (1 Viewer)

yashbb · Nov 15, 2021

Sorry it’s side ways

notme123 · Nov 15, 2021

cos50 + cos 70
= cos(60-10) + cos(60+10)
= cos60cos10 + sin60sin10 + cos60cos10 - sin60sin10
= 2cos60cos10
=2(1/2)cos10
=cos10
=sin(90-10)
=sin80

yashbb · Nov 15, 2021

notme123 said:
cos50 + cos 70
= cos(60-10) + cos(60+10)
= cos60cos10 + sin60sin10 + cos60cos10 - sin60sin10
= 2cos60cos10
=2(1/2)cos10
=cos10
=sin(90-10)
=sin80

tysmmm

CM_Tutor · Nov 15, 2021

Sums to products rule:

$\begin{align*} \cos\alpha + \cos\beta &= 2\cos{\left(\cfrac{\alpha + \beta}{2}\right)}\cos{\left(\cfrac{\alpha - \beta}{2}\right)} \\ \text{Put $\alpha = 50^\circ$ and $\beta = 70^\circ$:} \quad \cos{50^\circ} + \cos{70^\circ} &= 2\cos{\left(\cfrac{50^\circ + 70^\circ}{2}\right)}\cos{\left(\cfrac{50^\circ - 70^\circ}{2}\right)} \\ &= 2\cos{60^\circ}\cos{\left(-10^\circ\right)} \\ &= 2 \times \cfrac{1}{2} \times \cos{10^\circ} \qquad \text{as $\cos{60^\circ} = \cfrac{1}{2}$ and $\cos\theta$ is an even function} \\ &= 1 \times \sin{\left(90^\circ - 10^\circ\right)} \qquad \text{as $\cos\theta = \sin{\left(90^\circ - \theta\right)}$} \\ &= \sin{80^\circ} \end{align*}$

How do i part b (1 Viewer)

yashbb

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yashbb

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