Question

2(sin²75degrees +cos²75degrees)=​

Answer 1

Step-by-step explanation:

sin²x+cos²x=1

sin²75+cos²75=1

2(sin²75+cos²75)=2(1)=2

Answer 2

Solution :

$:\implies \sf{2(sin^{2}75^{\circ} + cos^{2}75^{\circ})}$

We know that, sin²75° and cos²75° can be written as sin²(30° + 45°) and cos²(30° + 45°), so substituting them in the equation, we get :

$:\implies \sf{2[sin^{2}(30^{\circ} + 45^{\circ}) + cos^{2}(30^{\circ} + 45^{\circ})]}$

Now by using the identity for cos(A + B) and sin(A + B) let's find out the value of sin²(30° + 45°) and cos²(30° + 45°), we get :

• $\sf{sin(A + B) = sinAcosB + cosAsinB}$

• $\sf{cos(A + B) = cosAcosB - sinAsinB}$

• Value of sin²(30° + 45°) :

By using the identity of sin(A + B), we get :

$:\implies \sf{sin^{2}(30^{\circ} + 45^{\circ}) = \bigg(sin30^{\circ}cos45^{\circ} + cos30^{\circ}sin45^{\circ}\bigg)^{2}}$

Now by substituting the values in sin30°, sin45°, cos30° and cos45°,i.e, 1/2 , 1/√2 , √3/2 and 1/√2, respectively in the equation, we get :

$:\implies \sf{sin^{2}(30^{\circ} + 45^{\circ}) = \bigg(\dfrac{1}{2} \times \dfrac{1}{\sqrt{2}} + \dfrac{\sqrt{3}}{2} \times \dfrac{1}{\sqrt{2}}\bigg)^{2}} \\ \\ :\implies \sf{sin^{2}(30^{\circ} + 45^{\circ}) = \bigg(\dfrac{1}{2\sqrt{2}} + \dfrac{\sqrt{3}}{2\sqrt{2}}\bigg)^{2}} \\ \\ :\implies \sf{sin^{2}(30^{\circ} + 45^{\circ}) = \bigg(\dfrac{1 + \sqrt{3}}{2\sqrt{2}}\bigg)^{2}}$

Now by using the identity of (a + b)², we get :

$:\implies \sf{sin^{2}(30^{\circ} + 45^{\circ}) = \dfrac{1^{2} + 2 \times \sqrt{3} + \sqrt{3}^{2}}{(2\sqrt{2})^{2}}} \\ \\ :\implies \sf{sin^{2}(30^{\circ} + 45^{\circ}) = \dfrac{1 + 2\sqrt{3} + 3}{8}} \\ \\ :\implies \sf{sin^{2}(30^{\circ} + 45^{\circ}) = \dfrac{4 + 2\sqrt{3}}{8}} \\ \\ \boxed{\therefore \sf{sin^{2}(30^{\circ} + 45^{\circ}) = \dfrac{4 + 2\sqrt{3}}{8}}}$

• Value of cos²(35° + 45°)

By using the identity of cos(A + B), we get :

$:\implies \sf{cos^{2}(30^{\circ} + 45^{\circ}) = \bigg(cos30^{\circ}cos45^{\circ} - sin30^{\circ}sin45^{\circ}\bigg)^{2}}$

Now by substituting the values in sin30°, sin45°, cos30° and cos45°,i.e, 1/2 , 1/√2 , √3/2 and 1/√2, respectively in the equation, we get :

$:\implies \sf{cos^{2}(30^{\circ} + 45^{\circ}) = \bigg(\dfrac{\sqrt{3}}{2} \times \dfrac{1}{\sqrt{2}} - \dfrac{1}{2} \times \dfrac{1}{\sqrt{2}}\bigg)^{2}} \\ \\ :\implies \sf{cos^{2}(30^{\circ} + 45^{\circ}) = \bigg(\dfrac{\sqrt{3}}{2\sqrt{2}} - \dfrac{1}{2\sqrt{2}}\bigg)^{2}} \\ \\ :\implies \sf{cos^{2}(30^{\circ} + 45^{\circ}) = \bigg(\dfrac{\sqrt{3} - 1}{2\sqrt{2}}\bigg)^{2}}$

Now by using the identity of (a - b)², we get :

$:\implies \sf{cos^{2}(30^{\circ} + 45^{\circ}) = \bigg(\dfrac{\sqrt{3} - 1}{2\sqrt{2}}\bigg)^{2}} \\ \\ :\implies \sf{cos^{2}(30^{\circ} + 45^{\circ}) = \dfrac{\sqrt{3}^{2} - 2 \times \sqrt{3} + 1^{2}}{(2\sqrt{2})^{2}}} \\ \\ :\implies \sf{cos^{2}(30^{\circ} + 45^{\circ}) = \dfrac{3 - 2\sqrt{3} + 1}{8}} \\ \\ :\implies \sf{cos^{2}(30^{\circ} + 45^{\circ}) = \dfrac{4 - 2\sqrt{3}}{8}} \\ \\ \boxed{\therefore \sf{sin^{2}(30^{\circ} + 45^{\circ}) = \dfrac{4 - 2\sqrt{3}}{8}}}$

Now by substituting the values of sin²(30° + 45) and cos²(30° + 45°) in the equation, we get :

$:\implies \sf{2(sin^{2}75^{\circ} + cos^{2}75^{\circ}) = 2\bigg(\dfrac{4 + 2\sqrt{3}}{8} + \dfrac{4 - 2\sqrt{3}}{8}\bigg)} \\ \\ :\implies \sf{2(sin^{2}75^{\circ} + cos^{2}75^{\circ}) = 2\bigg(\dfrac{4 + 2\sqrt{3} + 4 - 2\sqrt{3}}{8}\bigg)} \\ \\ :\implies \sf{2(sin^{2}75^{\circ} + cos^{2}75^{\circ}) = 2 \times \bigg(\dfrac{8}{8}\bigg)} \\ \\ :\implies \sf{2(sin^{2}75^{\circ} + cos^{2}75^{\circ}) = 2 \times 1} \\ \\ :\implies \sf{2(sin^{2}75^{\circ} + cos^{2}75^{\circ}) = 2} \\ \\ \boxed{\therefore \sf{2(sin^{2}75^{\circ} + cos^{2}75^{\circ}) = 2}}$