What is the value of cos(45^0)cos(60^0)+sin(45^0)sin(60^0) ?

The value of cos(45^0)cos(60^0)+sin(45^0)sin(60^0) is 1

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cos(45^0)cos(60^0)+sin(45^0)sin(60^0)

cos (4 5^{0}) cos (6 0^{0}) + sin (4 5^{0}) sin (6 0^{0})

1

Solution steps

cos (4 5^{0}) cos (6 0^{0}) + sin (4 5^{0}) sin (6 0^{0})

Simplify:

4 5^{0} = 1

4 5^{0}

Apply rule

a^{0} = 1, a \neq = 0

= 1

Simplify:

6 0^{0} = 1

6 0^{0}

Apply rule

a^{0} = 1, a \neq = 0

= 1

cos (1) cos (1) + sin (1) sin (1) = cos^{2} (1) + sin^{2} (1)

cos (1) cos (1) + sin (1) sin (1)

cos (1) cos (1) = cos^{2} (1)

cos (1) cos (1)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cos (1) cos (1) = cos^{1 + 1} (1)

= cos^{1 + 1} (1)

Add the numbers:

1 + 1 = 2

= cos^{2} (1)

sin (1) sin (1) = sin^{2} (1)

sin (1) sin (1)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

sin (1) sin (1) = sin^{1 + 1} (1)

= sin^{1 + 1} (1)

Add the numbers:

1 + 1 = 2

= sin^{2} (1)

= cos^{2} (1) + sin^{2} (1)

= cos^{2} (1) + sin^{2} (1)

Rewrite using trig identities:

1

cos^{2} (1) + sin^{2} (1)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

= 1

= 1

cos (15. 2^{\circ})

\frac{1}{2} sech^{2} (1) - \frac{1}{2} tanh (1)

3 cot (4 5^{\circ})

0.9144 + 3.048 sin (3 0^{\circ})

\frac{3 ( 2 ) ^{2} - sin ( 2 )}{2 e ^{2}}

What is the value of cos(45^0)cos(60^0)+sin(45^0)sin(60^0) ?
The value of cos(45^0)cos(60^0)+sin(45^0)sin(60^0) is 1