Question

MATH HELP PLZ!!!

Answer 1

Answer:

a)    $tan (157.5) = \frac{1-cos 315}{sin315}$

b)

            $sin (165) =\sqrt{ \frac{1-cos (330) }{2}}$

c)

      $sin^{2} (157.5) = \frac{1-cos (315) }{2}$

d)

  cos 330° = 1- 2 sin² (165°)

       

         

Step-by-step explanation:

Step(i):-

By using trigonometry formulas

a)

cos2∝  = 2 cos² ∝-1

cos∝ = 2 cos² ∝/2 -1

1+ cos∝ =  2 cos² ∝/2

$cos^{2} (\frac{\alpha }{2}) = \frac{1+cos\alpha }{2}$

b)

cos2∝  = 1- 2 sin² ∝

cos∝  = 1- 2 sin² ∝/2

$sin^{2} (\frac{\alpha }{2}) = \frac{1-cos\alpha }{2}$

Step(i):-

Given

              $tan\alpha = \frac{sin\alpha }{cos\alpha }$

          we know that trigonometry formulas

        $sin\alpha = 2sin(\frac{\alpha }{2} )cos(\frac{\alpha }{2} )$

         1- cos∝ =  2 sin² ∝/2

      Given

         $tan(\frac{\alpha }{2} ) = \frac{sin(\frac{\alpha }{2} )}{cos(\frac{\alpha }{2}) }$

put ∝ = 315

      $tan(\frac{315}{2} ) = \frac{sin(\frac{315 }{2} )}{cos(\frac{315 }{2}) }$

     multiply with ' 2 sin (∝/2) both numerator and denominator

        $tan (\frac{315}{2} )= \frac{2sin^{2}(\frac{315)}{2} }{2sin(\frac{315}{2} cos(\frac{315}{2}) }$

Apply formulas

 $sin\alpha = 2sin(\frac{\alpha }{2} )cos(\frac{\alpha }{2} )$

  1- cos∝ =  2 sin² ∝/2

now we get

 $tan (157.5) = \frac{1-cos 315}{sin315}$

       

b)

          $sin^{2} (\frac{\alpha }{2}) = \frac{1-cos\alpha }{2}$

               put ∝ = 330° above formula

             $sin^{2} (\frac{330 }{2}) = \frac{1-cos (330) }{2}$

            $sin^{2} (165) = \frac{1-cos (330) }{2}$

            $sin (165) =\sqrt{ \frac{1-cos (330) }{2}}$

c )

         $sin^{2} (\frac{\alpha }{2}) = \frac{1-cos\alpha }{2}$

               put ∝ = 315° above formula

             $sin^{2} (\frac{315 }{2}) = \frac{1-cos (315) }{2}$

            $sin^{2} (157.5) = \frac{1-cos (315) }{2}$

           

d)

     cos∝  = 1- 2 sin² ∝/2

   put      ∝ = 330°

       $cos 330 = 1 - 2sin^{2} (\frac{330}{2} )$

      cos 330° = 1- 2 sin² (165°)