Trigonometry-If-sin-A-sin-B-a-and-cos-A-cos-B-b-find-cos-A-B-in-terms-of-a-and-b-

Question Number 135895 by liberty last updated on 16/Mar/21

T r i g o n o m e t r y

If sin A + sin B = a and cos A + cos B = b , find cos (A+B) in terms of a and b?

Answered by EDWIN88 last updated on 18/Mar/21

{\begin{cases} \sin A + \sin B = 2 \sin (\frac{A + B}{2}) \cos (\frac{A - B}{2}) = a \\ \cos A + \cos B = 2 \cos (\frac{A + B}{2}) \cos (\frac{A - B}{2}) = b \end{cases}

we get \tan (\frac{A + B}{2}) = \frac{a}{b} and \tan (A + B) = \frac{2 \tan (\frac{A + B}{2})}{1 - \tan^{2} (\frac{A + B}{2})}

\tan (A + B) = \frac{(\frac{2 a}{b})}{1 - (\frac{a^{2}}{b^{2}})} = \frac{2 a b}{b^{2} - a^{2}}

then \sec^{2} (A + B) = 1 + \frac{4 a^{2} b^{2}}{{(b^{2} - a^{2})}^{2}} = \frac{{(b^{2} - a^{2})}^{2} + 4 a^{2} b^{2}}{{(b^{2} - a^{2})}^{2}}

s o \cos^{2} (A + B) = \frac{{(b^{2} - a^{2})}^{2}}{{(b^{2} + a^{2})}^{2}} \Rightarrow \cos (A + B) = \pm (\frac{b^{2} - a^{2}}{b^{2} + a^{2}})

Commented by mr W last updated on 17/Mar/21

p l e a s e c h e c k s i r :

i t h i n k ∣ \dots ∣ i s w r o n g . e x a m p l e

a = \sqrt{3}, b = 1 \Rightarrow \frac{b^{2} - a^{2}}{b^{2} + a^{2}} = - \frac{1}{2}

t h a t c o u l d m e a n A = \frac{π}{3}, B = \frac{π}{3}

\cos (A + B) = \cos (\frac{2 π}{3}) = - \frac{1}{2} = \frac{b^{2} - a^{2}}{b^{2} + a^{2}} \neq∣ \frac{b^{2} - a^{2}}{b^{2} + a^{2}} ∣

Commented by EDWIN88 last updated on 17/Mar/21

why sir . If x^{2} = {(b - a)}^{2} then x = \pm \sqrt{{(b - a)}^{2}}

x = ∣ b - a ∣ sir

Commented by mr W last updated on 17/Mar/21

x = \pm \sqrt{{(b - a)}^{2}} ≢ x = ∣ b - a ∣

w i t h x = ∣ b - a ∣ y o u f i x x > 0, b u t

x c a n a l s o b e n e g a t i v e : x = \pm ∣ b - a ∣

Answered by mr W last updated on 17/Mar/21

\sin (\frac{A + B}{2} + \frac{A - B}{2}) + \sin (\frac{A + B}{2} - \frac{A + B}{2}) = a

\sin \frac{A + B}{2} \cos \frac{A - B}{2} = \frac{a}{2} \dots (i)

\cos (\frac{A + B}{2} + \frac{A - B}{2}) + \cos (\frac{A + B}{2} - \frac{A + B}{2}) = b

\cos (\frac{A + B}{2}) \cos (\frac{A - B}{2}) = \frac{b}{2} \dots (i i)

(i) / (i i) :

\tan \frac{A + B}{2} = \frac{a}{b}

r e c a l l : \cos 2 α = \frac{1 - \tan^{2} α}{1 + \tan^{2} α}

\Rightarrow \cos (A + B) = \frac{1 - {(\frac{a}{b})}^{2}}{1 + {(\frac{a}{b})}^{2}} = \frac{b^{2} - a^{2}}{b^{2} + a^{2}}

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