Solveeit Logo
Login

Question

Question: If \(\cos \alpha + 2\cos \beta + 3\cos \lambda = 0,{\text{ }}\sin \alpha + 2\sin \beta + 3\sin \lamb...

If cos⁡α+2cos⁡β+3cos⁡λ=0, sin⁡α+2sin⁡β+3sin⁡λ=0 and α+β+λ=π,\cos \alpha + 2\cos \beta + 3\cos \lambda = 0,{\text{ }}\sin \alpha + 2\sin \beta + 3\sin \lambda = 0{\text{ and }}\alpha + \beta + \lambda = \pi , then
sin⁡3α+8sin⁡3β+27sin⁡3λ= a. - 18 b. 0 c. 3 d. 9  \sin 3\alpha + 8\sin 3\beta + 27\sin 3\lambda = \\\ {\text{a}}.{\text{ - 18}} \\\ {\text{b}}{\text{. 0}} \\\ {\text{c}}{\text{. 3}} \\\ {\text{d}}{\text{. 9}} \\\

Explanation

Solution

Hint: Assume z1=cos⁡α+isin⁡α, z2=cos⁡β+isin⁡β, z3=cos⁡λ+isin⁡λ{z_1} = \cos \alpha + i\sin \alpha ,{\text{ }}{{\text{z}}_2} = \cos \beta + i\sin \beta ,{\text{ }}{{\text{z}}_3} = \cos \lambda + i\sin \lambda . Manipulate these terms to obtain the required expression, also use Euler’s Theorem that is write complex numbers in exponential terms.

Complete step-by-step answer:
Let,
z1=cos⁡α+isin⁡α, z2=cos⁡β+isin⁡β, z3=cos⁡λ+isin⁡λ z1=cos⁡α+isin⁡α.........(1) 2z2=2(cos⁡β+isin⁡β)............(2) 3z3=3(cos⁡λ+isin⁡λ)...............(3)  {z_1} = \cos \alpha + i\sin \alpha ,{\text{ }}{{\text{z}}_2} = \cos \beta + i\sin \beta ,{\text{ }}{{\text{z}}_3} = \cos \lambda + i\sin \lambda \\\ {z_1} = \cos \alpha + i\sin \alpha .........\left( 1 \right) \\\ {\text{2}}{{\text{z}}_2} = 2\left( {\cos \beta + i\sin \beta } \right)............\left( 2 \right) \\\ {\text{3}}{{\text{z}}_3} = 3\left( {\cos \lambda + i\sin \lambda } \right)...............\left( 3 \right) \\\
Where zz is a complex number
Add these three equations
z1+2z2+3z3=cos⁡α+isin⁡α+2cos⁡β+2isin⁡β+3cos⁡λ+3isin⁡λ  =(cos⁡α+2cos⁡β+3cos⁡λ)+i(sin⁡α+2sin⁡β+3sin⁡λ)  {z_1} + 2{{\text{z}}_2} + 3{{\text{z}}_3} = \cos \alpha + i\sin \alpha + 2\cos \beta + 2i\sin \beta + 3\cos \lambda + 3i\sin \lambda \\\ {\text{ }} = \left( {\cos \alpha + 2\cos \beta + 3\cos \lambda } \right) + i\left( {\sin \alpha + 2\sin \beta + 3\sin \lambda } \right) \\\
Now it is given that cos⁡α+2cos⁡β+3cos⁡λ=0, sin⁡α+2sin⁡β+3sin⁡λ=0\cos \alpha + 2\cos \beta + 3\cos \lambda = 0,{\text{ }}\sin \alpha + 2\sin \beta + 3\sin \lambda = 0
⇒z1+2z2+3z3=0+i0=0\Rightarrow {z_1} + 2{{\text{z}}_2} + 3{{\text{z}}_3} = 0 + i0 = 0
Now according to Euler’s Theorem cos⁡α+isin⁡α=eiα\cos \alpha + i\sin \alpha = {e^{i\alpha }}
⇒z1=eiα ⇒z2=eiβ ⇒z3=eiλ  \Rightarrow {z_1} = {e^{i\alpha }} \\\ \Rightarrow {{\text{z}}_2} = {e^{i\beta }} \\\ \Rightarrow {{\text{z}}_3} = {e^{i\lambda }} \\\
Now according to known fact if a+b+c=0a + b + c = 0, then a3+b3+c3=3abc{a^3} + {b^3} + {c^3} = 3abc
⇒(z1)3+(2z2)3+(3z3)3=3(z1)(2z2)(3z3) ⇒e3iα+8e3iβ+27e3iλ=18eiαeiβeiλ ⇒cos⁡3α+isin⁡3α+8cos⁡3β+8isin⁡3β+27cos⁡3λ+27isin⁡3λ=18eiαeiβeiλ ⇒(cos⁡3α+8cos⁡3β+27cos⁡3λ)+i(sin⁡3α+8sin⁡3β+27sin⁡3λ)=18ei(α+β+λ)  \Rightarrow {\left( {{z_1}} \right)^3} + {\left( {2{{\text{z}}_2}} \right)^3} + {\left( {3{{\text{z}}_3}} \right)^3} = 3\left( {{z_1}} \right)\left( {2{{\text{z}}_2}} \right)\left( {3{{\text{z}}_3}} \right) \\\ \Rightarrow {e^{3i\alpha }} + 8{e^{3i\beta }} + 27{e^{3i\lambda }} = 18{e^{i\alpha }}{e^{i\beta }}{e^{i\lambda }} \\\ \Rightarrow \cos 3\alpha + i\sin 3\alpha + 8\cos 3\beta + 8i\sin 3\beta + 27\cos 3\lambda + 27i\sin 3\lambda = 18{e^{i\alpha }}{e^{i\beta }}{e^{i\lambda }} \\\ \Rightarrow \left( {\cos 3\alpha + 8\cos 3\beta + 27\cos 3\lambda } \right) + i\left( {\sin 3\alpha + 8\sin 3\beta + 27\sin 3\lambda } \right) = 18{e^{i\left( {\alpha + \beta + \lambda } \right)}} \\\
Now it is given that α+β+λ=π\alpha + \beta + \lambda = \pi
⇒(cos⁡3α+8cos⁡3β+27cos⁡3λ)+i(sin⁡3α+8sin⁡3β+27sin⁡3λ)=18ei(α+β+λ)=18eiπ ⇒(cos⁡3α+8cos⁡3β+27cos⁡3λ)+i(sin⁡3α+8sin⁡3β+27sin⁡3λ)=18(cos⁡π+isin⁡π)  \Rightarrow \left( {\cos 3\alpha + 8\cos 3\beta + 27\cos 3\lambda } \right) + i\left( {\sin 3\alpha + 8\sin 3\beta + 27\sin 3\lambda } \right) = 18{e^{i\left( {\alpha + \beta + \lambda } \right)}} = 18{e^{i\pi }} \\\ \Rightarrow \left( {\cos 3\alpha + 8\cos 3\beta + 27\cos 3\lambda } \right) + i\left( {\sin 3\alpha + 8\sin 3\beta + 27\sin 3\lambda } \right) = 18\left( {\cos \pi + i\sin \pi } \right) \\\
Now, as we know cos⁡π=−1, sin⁡π=0\cos \pi = - 1,{\text{ }}\sin \pi = 0
⇒(cos⁡3α+8cos⁡3β+27cos⁡3λ)+i(sin⁡3α+8sin⁡3β+27sin⁡3λ)=18(cos⁡π+isin⁡π)=18(−1+0i)=−18\Rightarrow \left( {\cos 3\alpha + 8\cos 3\beta + 27\cos 3\lambda } \right) + i\left( {\sin 3\alpha + 8\sin 3\beta + 27\sin 3\lambda } \right) = 18\left( {\cos \pi + i\sin \pi } \right) = 18\left( { - 1 + 0i} \right) = - 18Now comparing real and imaginary terms
cos⁡3α+8cos⁡3β+27cos⁡3λ=−18, sin⁡3α+8sin⁡3β+27sin⁡3λ=0  \cos 3\alpha + 8\cos 3\beta + 27\cos 3\lambda = - 18, \\\ \sin 3\alpha + 8\sin 3\beta + 27\sin 3\lambda = 0 \\\
So, the value of sin⁡3α+8sin⁡3β+27sin⁡3λ=0\sin 3\alpha + 8\sin 3\beta + 27\sin 3\lambda = 0
Hence, option b is correct.

Note: In such types of question always assume a complex number in the form of(z=cos⁡θ+isin⁡θ)\left( {z = \cos \theta + i\sin \theta } \right), then remember the standard known fact which is if a+b+c=0a + b + c = 0, then a3+b3+c3=3abc{a^3} + {b^3} + {c^3} = 3abc, then simplify according to given conditions then compare real and imaginary parts we will get the required answer.