Question

find the value of 2sin square 3π\4+ 2cos square 3π\4 - 2tan square 3π\4.​

Answer 1

Answer:

The value of the expression is zero.

Step-by-step explanation:

Given that 2sin²(3π/4)+2cos²(3π/4)-2tan²(3π/4)

We have to find the value of the above expression

$2sin^{2} (\frac{3\pi }{4} )+2cos^{2} (\frac{3\pi }{4} )-2tan^{2} (\frac{3\pi }{4} )$

We know that sin²A + cos²A = 1

tan(135°) = tan(180° - 45°) = -tan(45°) = -1

tan²(135°) = (-1)² = 1

tangent function is negative in the second quadrant.

$2sin^{2} (\frac{3\pi }{4} )+2cos^{2} (\frac{3\pi }{4} )-2tan^{2} (\frac{3\pi }{4} )$

$=2(sin^{2} (\frac{3\pi }{4} )+cos^{2} (\frac{3\pi }{4} ))-2tan^{2} (\frac{3\pi }{4} )$

=$2(1)-2tan^{2} (\frac{3\pi }{4} )$

$=2(1-tan^{2}( \frac{3\pi }{4} ))$

$=2(1-1)$

=2(0)

=0

Therefore,  $2sin^{2} (\frac{3\pi }{4} )+2cos^{2} (\frac{3\pi }{4} )-2tan^{2} (\frac{3\pi }{4} )$ = 0

Answer 2

Answer:

The value of the given equation is 0.

Step-by-step explanation:

Given trigonometric equation is:

$\[2{\sin ^2}\left( {\frac{{3\pi }}{4}} \right) + 2{\cos ^2}\left( {\frac{{3\pi }}{4}} \right) - 2{\tan ^2}\left( {\frac{{3\pi }}{4}} \right)\]$

We will split the angle in the form of $\[\pi - \theta \]$ to simplify the given equation.

$\[ \Rightarrow 2{\sin ^2}\left( {\pi - \frac{\pi }{4}} \right) + 2{\cos ^2}\left( {\pi - \frac{\pi }{4}} \right) - 2{\tan ^2}\left( {\pi - \frac{\pi }{4}} \right)\]$

As we know that,

sin(π–θ) = sin(θ)

cos(π–θ) = –cos(θ)

tan(π–θ) = –tan(θ)

Therefoer,

$\[\begin{gathered} \Rightarrow 2{\left( {\sin \frac{\pi }{4}} \right)^2} + 2{\left( { - \cos \frac{\pi }{4}} \right)^2} - 2{\left( { - \tan \frac{\pi }{4}} \right)^2} \hfill \\ \Rightarrow 2{\left( {\sin \frac{\pi }{4}} \right)^2} + 2{\left( { - \cos \frac{\pi }{4}} \right)^2} - 2{\left( { - \tan \frac{\pi }{4}} \right)^2} \hfill \\ \end{gathered} \]$

Since the values of

sin$\frac{\pi}{4}$ = $\frac{1}{\sqrt{2} }$

cos$\frac{\pi}{4}$ = $\frac{1}{\sqrt{2} }$

tan$\frac{\pi}{4}$ = 1

Therefore,

$\[\begin{gathered} \Rightarrow 2{\left( {\frac{1}{{\sqrt 2 }}} \right)^2} + 2{\left( { - \frac{1}{{\sqrt 2 }}} \right)^2} - 2{\left( { - 1} \right)^2} \hfill \\ \Rightarrow 2 \times \frac{1}{2} + 2 \times \frac{1}{2} - 2 \hfill \\ \Rightarrow 1 + 1 - 2 \hfill \\ \Rightarrow 2 - 2 \hfill \\ \Rightarrow 0 \hfill \\ \end{gathered} \]$

Therefore, the value of the given equation is 0.