Question

$\bf \pmb{if \: a + b + c = \frac{\pi}{2} } \\ \bf \pmb{then \: find \: \sin(2a) + \sin(2b) + \sin(2c) }$
solve above question on trigonometry​

Answer 1

Answer:

4cosa.cosb.cosc

Step-by-step explanation:

Given :

$\mathrm{a + b + c =\dfrac{\pi}{2}}$

To find :

sin(2a) + sin(2b) + sin(2c)

Solution :

=> sin(2a) + sin(2b) + sin(2c)

We know that,

$\boxed{\mathrm{sinC + sinD = 2sin\bigg(\dfrac{C+D}{2}\bigg)cos\bigg(\dfrac{C-D}{2}\bigg)}}$

here, C = 2a and D =2b

So, applying for 1st 2 terms,

$\implies \mathrm{2sin\bigg(\dfrac{2a + 2b}{2}\bigg)cos\bigg(\dfrac{2a-2b}{2}\bigg) + sin2c}$

$\implies \mathrm{2sin(a+b)cos(a-b)+sin2c}$

From given,

$\mathrm{a + b + c =\dfrac{\pi}{2}}$

So,

$\mathrm{a + b =\dfrac{\pi}{2} -c}$

Substituting in place of a + b, we get

$\implies \mathrm{2sin\bigg(\dfrac{\pi}{2}-c\bigg)cos(a-b)+sin2c}$

We know that,

$\boxed{\mathrm{sin\bigg(\dfrac{\pi}{2}-x\bigg) = cosx}}$

So,

$\implies \mathrm{2cosc\; cos(a-b)+sin2c}$

We know that,

$\boxed{\mathrm{sin2x = 2sinx\,cox}}$

$\implies \mathrm{2cosc\; cos(a-b)+2sinc\,cosc}$

Taking 2cosc common,

$\implies \mathrm{2cosc(cos(a-b)+sinc)}$

From given,

$\mathrm{a + b + c =\dfrac{\pi}{2}}$

So,

$\mathrm{c =\dfrac{\pi}{2} -(a+b)}$

$\implies \mathrm{2cosc(cos(a-b)+sin\bigg(\dfrac{\pi}{2} - (a+b)\bigg)}$

We know that,

$\boxed{\mathrm{sin\bigg(\dfrac{\pi}{2}-x\bigg) = cosx}}$

$\implies \mathrm{2cosc(cos(a-b)+cos(a+b))}$

We know that,

$\boxed{\mathrm{cosC + cosD = 2cos\bigg(\dfrac{C+D}{2}\bigg)cos\bigg(\dfrac{C-D}{2}\bigg)}}$

So,

$\implies \mathrm{2cosc\;2cos\bigg(\dfrac{a+b+a-b}{2}\bigg) cos\bigg(\dfrac{a-b-a-b}{2}\bigg)}$

$\implies \mathrm{4 cosa\; cosb\; cosc}$

$\therefore \overline{\underline{\boxed{\begin{array}{cc}\textbf{if a+b+c = \dfrac{\pi}{\bf 2} }\\\textbf{then}\\\mathbf{sin2a+sin2b+sin2c = 4cosa\,cosb\,cosc}\end{array}}}}$

❖ Extra information

⟡ Transformation formulae :

$\boxed{\begin{array}{cc} \boxed{\mathrm{sinC + sinD = 2sin\bigg(\dfrac{C+D}{2}\bigg)cos\bigg(\dfrac{C-D}{2}\bigg)}}\\\\\boxed{\mathrm{sinC - sinD = 2cos\bigg(\dfrac{C+D}{2}\bigg)sin\bigg(\dfrac{C-D}{2}\bigg)}}\\\\\boxed{\mathrm{cosC + cosD = 2cos\bigg(\dfrac{C+D}{2}\bigg)cos\bigg(\dfrac{C-D}{2}\bigg)}}\\\\\boxed{\mathrm{cosC - cosD = -2sin\bigg(\dfrac{C+D}{2}\bigg)sin\bigg(\dfrac{C-D}{2}\bigg)}}\end{array}}$

Hope it helps!