Question-161079

Question Number 161079 by 0731619 last updated on 11/Dec/21

Commented by mr W last updated on 11/Dec/21

must >0 and <1, therefore only (1/(16)) can be, if one answer should be true.

$${must}\:>\mathrm{0}\:{and}\:<\mathrm{1},\:{therefore}\:{only}\:\frac{\mathrm{1}}{\mathrm{16}} \\ $$$${can}\:{be},\:{if}\:{one}\:{answer}\:{should}\:{be}\:{true}. \\ $$

Commented by 0731619 last updated on 11/Dec/21

$${solution}\:{sir}? \\ $$

Commented by infinityaction last updated on 23/Apr/22

use formula (1/4)cos 3x =cos (60−x)cos (x)cos (60+x) put x=25 (1/4)cos 75 = cos 35cos 25cos 85 p = cos 15((1/4)cos 75) p = (2/8)cos 15sin 15 p = (1/8)×sin 30 p = (1/(16))

$${use}\:{formula} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\mathrm{cos}\:\mathrm{3}{x}\:=\mathrm{cos}\:\left(\mathrm{60}−{x}\right)\mathrm{cos}\:\left({x}\right)\mathrm{cos}\:\left(\mathrm{60}+{x}\right) \\ $$$$\:{put}\:{x}=\mathrm{25} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\mathrm{cos}\:\mathrm{75}\:=\:\mathrm{cos}\:\mathrm{35cos}\:\mathrm{25cos}\:\mathrm{85} \\ $$$${p}\:=\:\mathrm{cos}\:\mathrm{15}\left(\frac{\mathrm{1}}{\mathrm{4}}\mathrm{cos}\:\mathrm{75}\right) \\ $$$${p}\:=\:\frac{\mathrm{2}}{\mathrm{8}}\mathrm{cos}\:\mathrm{15sin}\:\mathrm{15} \\ $$$${p}\:=\:\frac{\mathrm{1}}{\mathrm{8}}×\mathrm{sin}\:\mathrm{30} \\ $$$${p}\:=\:\frac{\mathrm{1}}{\mathrm{16}} \\ $$$$ \\ $$$$ \\ $$

Answered by TheSupreme last updated on 11/Dec/21

(1/4)[cos40+cos(10)][cos(50)+cos(120)]= =(1/4)[cos(40)cos(50)+cos(10)cos(50)−(1/2)cos(40)−(1/2)cos(10)] =(1/4)[(1/2)(cos(90)+cos(10))+(1/2)(cos(60)+cos(40))−(1/2)(cos(10)+cos(40)] (1/4)[(1/2)cos(60)]=(1/(16))

$$\frac{\mathrm{1}}{\mathrm{4}}\left[{cos}\mathrm{40}+{cos}\left(\mathrm{10}\right)\right]\left[{cos}\left(\mathrm{50}\right)+{cos}\left(\mathrm{120}\right)\right]= \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left[{cos}\left(\mathrm{40}\right){cos}\left(\mathrm{50}\right)+{cos}\left(\mathrm{10}\right){cos}\left(\mathrm{50}\right)−\frac{\mathrm{1}}{\mathrm{2}}{cos}\left(\mathrm{40}\right)−\frac{\mathrm{1}}{\mathrm{2}}{cos}\left(\mathrm{10}\right)\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left[\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\left(\mathrm{90}\right)+{cos}\left(\mathrm{10}\right)\right)+\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\left(\mathrm{60}\right)+{cos}\left(\mathrm{40}\right)\right)−\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\left(\mathrm{10}\right)+{cos}\left(\mathrm{40}\right)\right]\right. \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left[\frac{\mathrm{1}}{\mathrm{2}}{cos}\left(\mathrm{60}\right)\right]=\frac{\mathrm{1}}{\mathrm{16}} \\ $$

Commented by mr W last updated on 12/Dec/21

$${good}\:{sir}!\:{thanks}! \\ $$

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