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Question Number 60533 by Tawa1 last updated on 21/May/19
If A, B, C are angle of a triangle. Show that cos (1/2)C + cos (1/2)(A − B) = 2 sin (1/2)A sin (1/2)B
$$\mathrm{If}\:\:\mathrm{A},\:\mathrm{B},\:\mathrm{C}\:\:\mathrm{are}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}.\:\mathrm{Show}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{C}\:+\:\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{A}\:−\:\mathrm{B}\right)\:\:=\:\:\mathrm{2}\:\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{A}\:\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{B} \\ $$
Commented by malwaan last updated on 21/May/19
A+B+C=180^° cos(x+y)−cos(x−y)=−2sin(x)sin(y) ∴ 2sin(x)sin(y)=cos(x−y)−cos(x+y) R.H.S. 2sin(1/2)Asin(1/2)B=cos(1/2)(A−B)−cos(1/2)(A+B) =cos(1/2)(A−B)−cos(1/2)(180−C) =cos(1/2)(A−B)−cos(90−(C/2)) =cos(1/2)(A−B)−sin(1/2)C ≠L.H.S. whats wrong? the question or me?
$$\mathrm{A}+\mathrm{B}+\mathrm{C}=\mathrm{180}^{°} \\ $$$${cos}\left({x}+{y}\right)−{cos}\left({x}−{y}\right)=−\mathrm{2}{sin}\left({x}\right){sin}\left({y}\right) \\ $$$$\therefore\:\mathrm{2}{sin}\left({x}\right){sin}\left({y}\right)={cos}\left({x}−{y}\right)−{cos}\left({x}+{y}\right) \\ $$$${R}.{H}.{S}. \\ $$$$\mathrm{2}{sin}\frac{\mathrm{1}}{\mathrm{2}}{Asin}\frac{\mathrm{1}}{\mathrm{2}}{B}={cos}\frac{\mathrm{1}}{\mathrm{2}}\left({A}−{B}\right)−{cos}\frac{\mathrm{1}}{\mathrm{2}}\left({A}+{B}\right) \\ $$$$={cos}\frac{\mathrm{1}}{\mathrm{2}}\left({A}−{B}\right)−{cos}\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{180}−{C}\right) \\ $$$$={cos}\frac{\mathrm{1}}{\mathrm{2}}\left({A}−{B}\right)−{cos}\left(\mathrm{90}−\frac{{C}}{\mathrm{2}}\right) \\ $$$$={cos}\frac{\mathrm{1}}{\mathrm{2}}\left({A}−{B}\right)−{sin}\frac{\mathrm{1}}{\mathrm{2}}{C} \\ $$$$\neq\mathrm{L}.\mathrm{H}.\mathrm{S}. \\ $$$${whats}\:{wrong}? \\ $$$${the}\:{question}\:{or}\:{me}? \\ $$
Commented by Tawa1 last updated on 21/May/19
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by MJS last updated on 21/May/19
let A=60° B=45° C=75° cos 37.5° +cos 7.5° ≈1.78 2sin 30° sin 22.5° ≈0.38 formula is wrong
$$\mathrm{let}\:{A}=\mathrm{60}°\:{B}=\mathrm{45}°\:{C}=\mathrm{75}° \\ $$$$\mathrm{cos}\:\mathrm{37}.\mathrm{5}°\:+\mathrm{cos}\:\mathrm{7}.\mathrm{5}°\:\approx\mathrm{1}.\mathrm{78} \\ $$$$\mathrm{2sin}\:\mathrm{30}°\:\mathrm{sin}\:\mathrm{22}.\mathrm{5}°\:\approx\mathrm{0}.\mathrm{38} \\ $$$$\mathrm{formula}\:\mathrm{is}\:\mathrm{wrong} \\ $$
Commented by Tawa1 last updated on 21/May/19
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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