Question

(sin90°+cos60°+cos45°)×(sin30°+cos0°-cos45°)​

Answer 1

Given: (sin90°+cos60°+cos45°)×(sin30°+cos0°-cos45°)

As we know that,

                              Sin90°= 1

                             Cos60°= 1/2

                              Cos45°= 1/$\sqrt{2}$

                              Sin30°= 1/2

                              Cos0°= 1

putting all the values in given equation,

             = $(1+\frac{1}{2} +\frac{1}{\sqrt{2} })$×$(\frac{1}{1} +1-\frac{1}{\sqrt{2} } )$

             = $(\frac{\sqrt[2]{2}+\sqrt{2} +2 }{\sqrt[2]{2} })$×$(\frac{\sqrt{2}+\sqrt[2]{2}-2 }{\sqrt[2]{2} } )$

             =$(\frac{\sqrt[3]{2}+2 }{\sqrt[2]{2} } )$×$(\frac{\sqrt[3]{2}-2 }{\sqrt[2]{2} } )$

             =$\frac{(\sqrt[3]{2}) ^{2}-2^{2} }{8}$

             = $\frac{9(2)-4}{8}$

             = $\frac{18-4}{8}$

             = $\frac{14}{8}$

             =$\frac{7}{4}$

  Hence the final value of the given equation is 7/4.

Answer 2

Answer:

Solutions

(cos0

o

+sin45

o

+sin30

o

)(sin90

o

−cos45

o

+cos60

o

)

=(1+

2

1

+

2

1

)(1−

2

1

+

2

1

)

=(1+

2

1

+

2

1

)(1+

2

1

−

2

1

)

=(

2

3

+

2

1

)(

2

3

−

2

1

)

=(

2

3

)

2

−(

2

1

)

2

=

4

9

−

2

1

=

4

7