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Question Number 6145 by Ninik last updated on 16/Jun/16
show that sin 64^(° )  ×cos  26^°  + cos 64^(° ) × sin 26^°  =1
$${show}\:{that}\:\mathrm{sin}\:\mathrm{64}^{°\:} \:×\mathrm{cos}\:\:\mathrm{26}^{°} \:+\:\mathrm{cos}\:\mathrm{64}^{°\:} ×\:\mathrm{sin}\:\mathrm{26}^{°} \:=\mathrm{1}\: \\ $$$$ \\ $$
Answered by Rasheed Soomro last updated on 16/Jun/16
 sin 64^(° )  ×cos  26^°  + cos 64^(° ) × sin 26^°  =1   LHS=sin(64+26)=sin 90=1=RHS  Formula used:          sin α cos β+cos α sin β=sin (α+β)
$$\:\mathrm{sin}\:\mathrm{64}^{°\:} \:×\mathrm{cos}\:\:\mathrm{26}^{°} \:+\:\mathrm{cos}\:\mathrm{64}^{°\:} ×\:\mathrm{sin}\:\mathrm{26}^{°} \:=\mathrm{1}\: \\ $$$${LHS}=\mathrm{sin}\left(\mathrm{64}+\mathrm{26}\right)=\mathrm{sin}\:\mathrm{90}=\mathrm{1}={RHS} \\ $$$${Formula}\:{used}: \\ $$$$\:\:\:\:\:\:\:\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\beta+\mathrm{cos}\:\alpha\:\mathrm{sin}\:\beta=\mathrm{sin}\:\left(\alpha+\beta\right) \\ $$

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