Question

sin go= sin o-Sin2 45X cos² 45+ Seco x coso 2 ​

Answer 1

Answer:

Given expression cos

2

(45

∘

+x)+(sinx−cosx)

2

=

2

1

(1+cos(90+2x))+(sinx−cosx)

2

[∵cos2x=2cos

2

x−1]

=

2

1

(1−sin2x)+(sin

2

x−2sinxcosx+cos

2

x)

=

2

1

(1−sin2x)+1−2sinxcosx

=

2

1

(1−sin2x)+1−sin2x

=

2

3

(1−sin2x)

Now, since −1≤sin2x≤1

−1≤−sin2x≤1

⇒0≤1−sin2x≤2

⇒0≤

2

3

(1−sin2x)≤3

Hence,maximum value is 3

Step-by-step explanation:

hope helps