Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 121689 by bemath last updated on 11/Nov/20

Let a and b be real numbers such that { ((sin a + sin b = ((√2)/2))),((cos a + cos b = ((√6)/2))) :} Evaluate sin (a+b).

$$\:{Let}\:{a}\:{and}\:{b}\:{be}\:{real}\:{numbers}\:{such} \\ $$$${that}\:\begin{cases}{\mathrm{sin}\:{a}\:+\:\mathrm{sin}\:{b}\:=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\\{\mathrm{cos}\:{a}\:+\:\mathrm{cos}\:{b}\:=\:\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}}\end{cases} \\ $$$${Evaluate}\:\mathrm{sin}\:\left({a}+{b}\right).\: \\ $$

Answered by liberty last updated on 11/Nov/20

squaring the both equation gives { ((sin^2 a + 2sin a sin b + sin^2 b = (2/4))),((cos^2 a +2 cos a cos b + cos^2 b = (6/4))) :} and add the result to obtain 2 + 2 cos (a−b) = 2 → cos (a−b) = 0 Multiplying the two given relation yields ⇒(sin a + sin b)(cos a+cos b) = ((√3)/2) ⇒sin acos a + sin acos b+sin bcos a+sin bcos b=((√3)/2) ⇒sin 2a + sin 2b+2sin (a+b)= (√3) ⇒2sin (a+b)cos (a−b)+2sin (a+b)=(√3) ⇒2sin (a+b) { cos (a−b)+1 } = (√3) ⇒2sin (a+b) { 0+1 } = (√3) ⇒sin (a+b) = ((√3)/2). ▲

$$\mathrm{squaring}\:\mathrm{the}\:\mathrm{both}\:\mathrm{equation}\:\mathrm{gives} \\ $$$$\begin{cases}{\mathrm{sin}\:^{\mathrm{2}} {a}\:+\:\mathrm{2sin}\:{a}\:\mathrm{sin}\:{b}\:+\:\mathrm{sin}\:^{\mathrm{2}} {b}\:=\:\frac{\mathrm{2}}{\mathrm{4}}}\\{\mathrm{cos}\:^{\mathrm{2}} {a}\:+\mathrm{2}\:\mathrm{cos}\:{a}\:\mathrm{cos}\:{b}\:+\:\mathrm{cos}\:^{\mathrm{2}} {b}\:=\:\frac{\mathrm{6}}{\mathrm{4}}}\end{cases} \\ $$$$\mathrm{and}\:\mathrm{add}\:\mathrm{the}\:\mathrm{result}\:\mathrm{to}\:\mathrm{obtain}\: \\ $$$$\mathrm{2}\:+\:\mathrm{2}\:\mathrm{cos}\:\left({a}−{b}\right)\:=\:\mathrm{2}\:\rightarrow\:\mathrm{cos}\:\left({a}−{b}\right)\:=\:\mathrm{0} \\ $$$$\mathrm{Multiplying}\:\mathrm{the}\:\mathrm{two}\:\mathrm{given}\:\mathrm{relation}\:\mathrm{yields} \\ $$$$\Rightarrow\left(\mathrm{sin}\:{a}\:+\:\mathrm{sin}\:{b}\right)\left(\mathrm{cos}\:{a}+\mathrm{cos}\:{b}\right)\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{sin}\:{a}\mathrm{cos}\:{a}\:+\:\mathrm{sin}\:{a}\mathrm{cos}\:{b}+\mathrm{sin}\:{b}\mathrm{cos}\:{a}+\mathrm{sin}\:{b}\mathrm{cos}\:{b}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{sin}\:\mathrm{2}{a}\:+\:\mathrm{sin}\:\mathrm{2}{b}+\mathrm{2sin}\:\left({a}+{b}\right)=\:\sqrt{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{2sin}\:\left({a}+{b}\right)\mathrm{cos}\:\left({a}−{b}\right)+\mathrm{2sin}\:\left({a}+{b}\right)=\sqrt{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{2sin}\:\left({a}+{b}\right)\:\left\{\:\mathrm{cos}\:\left({a}−{b}\right)+\mathrm{1}\:\right\}\:=\:\sqrt{\mathrm{3}}\: \\ $$$$\Rightarrow\mathrm{2sin}\:\left({a}+{b}\right)\:\left\{\:\mathrm{0}+\mathrm{1}\:\right\}\:=\:\sqrt{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{sin}\:\left({a}+{b}\right)\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}.\:\blacktriangle \\ $$

Answered by TANMAY PANACEA last updated on 11/Nov/20

2sin(((a+b)/2))cos(((a−b)/2))=((√2)/2) 2cos(((a+b)/2))cos(((a−b)/2))=((√6)/2) tan(((a+b)/2))=((√2)/( (√6)))=(1/( (√3)))=tan30^o a+b=60^o sin(a+b)=((√3)/2) Or method or sin(a+b)=((2t)/(1+t^2 )) [t=tan((a+b)/2)] =((2×(1/( (√3))))/(1+((1/( (√3))))^2 ))=((2/( (√3)))/(4/3))=((2×3)/(4×(√3)))=((√3)/2)

$$\mathrm{2}{sin}\left(\frac{{a}+{b}}{\mathrm{2}}\right){cos}\left(\frac{{a}−{b}}{\mathrm{2}}\right)=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\mathrm{2}{cos}\left(\frac{{a}+{b}}{\mathrm{2}}\right){cos}\left(\frac{{a}−{b}}{\mathrm{2}}\right)=\frac{\sqrt{\mathrm{6}}}{\mathrm{2}} \\ $$$${tan}\left(\frac{{a}+{b}}{\mathrm{2}}\right)=\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{6}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}={tan}\mathrm{30}^{{o}} \\ $$$${a}+{b}=\mathrm{60}^{{o}} \\ $$$${sin}\left({a}+{b}\right)=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\boldsymbol{{O}}{r}\:{method} \\ $$$${or}\:{sin}\left({a}+{b}\right)=\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\:\:\left[{t}={tan}\frac{{a}+{b}}{\mathrm{2}}\right] \\ $$$$=\frac{\mathrm{2}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}{\mathrm{1}+\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} }=\frac{\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}}{\frac{\mathrm{4}}{\mathrm{3}}}=\frac{\mathrm{2}×\mathrm{3}}{\mathrm{4}×\sqrt{\mathrm{3}}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com