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Friday, August 6, 2021

2021/057) Given a^2+b^2=c^2+d^2=1 and ac+bd=0.Compute the value of ab+cd

Because a^2 + b^2 = 1 we can take a = \sin\alpha and b = \cos \alpha

Similarly c=\sin \beta and d = \cos\beta

ac + bd = \sin\alpha \sin  \beta + \cos \alpha \cos \beta = 0

or \cos (\alpha - \beta) = 0\cdots (1)  (using formula of cos of difference)

ab + cd = \sin \alpha \cos \alpha + \sin \beta \cos \beta = \frac{1}{2}(\sin 2 \alpha + \sin  2 \beta) (using formula for sin of twice angle)  

= \frac{1}{2} \sin (\alpha + \beta) \cos (\alpha - \beta) using sum of sines

= \frac{1}{2} \sin (\alpha + \beta) * 0 using (1)

= 0 

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