If-sin-x-cos-x-1-3-then-sin-3-x-cos-3-x-

Question Number 127944 by liberty last updated on 03/Jan/21

$$\:\mathrm{If}\:\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{then}\:\mathrm{sin}\:^{\mathrm{3}} \mathrm{x}\:+\:\mathrm{cos}\:^{\mathrm{3}} \mathrm{x}\:=? \\ $$

Answered by mr W last updated on 03/Jan/21

p_n =sin^n x+cos^n x p_1 =(1/3)=e_1 p_2 =e_1 ^2 −2e_2 =(1/9)−2e_2 =1 ⇒e_2 =−(4/9) p_3 =e_1 ^3 −3e_1 e_2 =(1/(27))+3×(1/3)×(4/9)=((13)/(27)) i.e. sin^3 x + cos^3 x =((13)/(27))

$${p}_{{n}} =\mathrm{sin}^{{n}} \:{x}+\mathrm{cos}^{{n}} \:{x} \\ $$$${p}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{3}}={e}_{\mathrm{1}} \\ $$$${p}_{\mathrm{2}} ={e}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{2}{e}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{9}}−\mathrm{2}{e}_{\mathrm{2}} =\mathrm{1}\:\Rightarrow{e}_{\mathrm{2}} =−\frac{\mathrm{4}}{\mathrm{9}} \\ $$$${p}_{\mathrm{3}} ={e}_{\mathrm{1}} ^{\mathrm{3}} −\mathrm{3}{e}_{\mathrm{1}} {e}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{27}}+\mathrm{3}×\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{4}}{\mathrm{9}}=\frac{\mathrm{13}}{\mathrm{27}} \\ $$$${i}.{e}.\:\mathrm{sin}\:^{\mathrm{3}} \mathrm{x}\:+\:\mathrm{cos}\:^{\mathrm{3}} \mathrm{x}\:=\frac{\mathrm{13}}{\mathrm{27}} \\ $$

Commented by bramlexs22 last updated on 04/Jan/21

if p_n = sin^n (x)−cos^n (x) how to get sin^3 (x)+cos^3 (x) sir?

$$\mathrm{if}\:\mathrm{p}_{\mathrm{n}} \:=\:\mathrm{sin}\:^{\mathrm{n}} \left(\mathrm{x}\right)−\mathrm{cos}\:^{\mathrm{n}} \left(\mathrm{x}\right) \\ $$$$\mathrm{how}\:\mathrm{to}\:\mathrm{get}\:\mathrm{sin}\:^{\mathrm{3}} \left(\mathrm{x}\right)+\mathrm{cos}\:^{\mathrm{3}} \left(\mathrm{x}\right)\:\mathrm{sir}? \\ $$

Commented by mr W last updated on 04/Jan/21

with p_n = sin^n (x)−cos^n (x) you get using this method corespondingly sin^3 (x)−cos^3 (x).

$${with}\:\:\mathrm{p}_{\mathrm{n}} \:=\:\mathrm{sin}\:^{\mathrm{n}} \left(\mathrm{x}\right)−\mathrm{cos}\:^{\mathrm{n}} \left(\mathrm{x}\right)\:{you}\:{get} \\ $$$${using}\:{this}\:{method}\:{corespondingly} \\ $$$$\:\mathrm{sin}\:^{\mathrm{3}} \left(\mathrm{x}\right)−\mathrm{cos}\:^{\mathrm{3}} \left(\mathrm{x}\right). \\ $$

Answered by bramlexs22 last updated on 03/Jan/21

Commented by mr W last updated on 03/Jan/21

in question: sin x+cos x=(1/3), not sin x−cos x=(1/3).

$${in}\:{question}:\:\mathrm{sin}\:{x}+\mathrm{cos}\:{x}=\frac{\mathrm{1}}{\mathrm{3}},\:{not} \\ $$$$\mathrm{sin}\:{x}−\mathrm{cos}\:{x}=\frac{\mathrm{1}}{\mathrm{3}}. \\ $$

Commented by bramlexs22 last updated on 03/Jan/21

�� misread...

Commented by liberty last updated on 03/Jan/21

$$\mathrm{hahahaha} \\ $$

Answered by mathmax by abdo last updated on 03/Jan/21

sin^3 x +cos^3 x =(sinx +cosx)(sin^2 x−sinx cosx +cos^2 x) =(1/3)(1−sinx cosx) we have (cosx +sinx)^2 =(1/9) ⇒ 1+2cosx sinx =(1/9) ⇒2cosx sinx=(1/9)−1 =−(8/9) ⇒cosx sinx=−(4/9) ⇒cos^3 x +sin^3 x =(1/3)(1+(4/9))=(1/3)×((13)/9) =((13)/(27))

$$\mathrm{sin}^{\mathrm{3}} \mathrm{x}\:+\mathrm{cos}^{\mathrm{3}} \mathrm{x}\:=\left(\mathrm{sinx}\:+\mathrm{cosx}\right)\left(\mathrm{sin}^{\mathrm{2}} \mathrm{x}−\mathrm{sinx}\:\mathrm{cosx}\:+\mathrm{cos}^{\mathrm{2}} \mathrm{x}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{1}−\mathrm{sinx}\:\mathrm{cosx}\right)\:\:\mathrm{we}\:\mathrm{have}\:\left(\mathrm{cosx}\:+\mathrm{sinx}\right)^{\mathrm{2}} \:=\frac{\mathrm{1}}{\mathrm{9}}\:\Rightarrow \\ $$$$\mathrm{1}+\mathrm{2cosx}\:\mathrm{sinx}\:=\frac{\mathrm{1}}{\mathrm{9}}\:\Rightarrow\mathrm{2cosx}\:\mathrm{sinx}=\frac{\mathrm{1}}{\mathrm{9}}−\mathrm{1}\:=−\frac{\mathrm{8}}{\mathrm{9}}\:\Rightarrow\mathrm{cosx}\:\mathrm{sinx}=−\frac{\mathrm{4}}{\mathrm{9}} \\ $$$$\Rightarrow\mathrm{cos}^{\mathrm{3}} \mathrm{x}\:+\mathrm{sin}^{\mathrm{3}} \mathrm{x}\:=\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{1}+\frac{\mathrm{4}}{\mathrm{9}}\right)=\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{13}}{\mathrm{9}}\:=\frac{\mathrm{13}}{\mathrm{27}} \\ $$

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