Is sqrt(2)sin(x+45)=sin(x)+cos(x) ?

The answer to whether sqrt(2)sin(x+45)=sin(x)+cos(x) is True

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Popular Trigonometry >

prove sqrt(2)sin(x+45)=sin(x)+cos(x)

Solution

prove

2 sin (x + 4 5^{\circ}) = sin (x) + cos (x)

Solution

True

Solution steps

2 sin (x + 4 5^{\circ}) = sin (x) + cos (x)

Manipulating left side

2 sin (x + 4 5^{\circ})

Rewrite using trig identities

sin (x + 4 5^{\circ})

Use the Angle Sum identity:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= sin (x) cos (4 5^{\circ}) + cos (x) sin (4 5^{\circ})

Simplify

sin (x) cos (4 5^{\circ}) + cos (x) sin (4 5^{\circ}) : \frac{2 sin ( x ) + 2 cos ( x )}{2}

sin (x) cos (4 5^{\circ}) + cos (x) sin (4 5^{\circ})

sin (x) cos (4 5^{\circ}) = \frac{2 sin ( x )}{2}

sin (x) cos (4 5^{\circ})

Simplify

cos (4 5^{\circ}) : \frac{2}{2}

cos (4 5^{\circ})

Use the following trivial identity:

cos (4 5^{\circ}) = \frac{2}{2}

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{2}{2}

= \frac{2}{2} sin (x)

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{2 sin ( x )}{2}

cos (x) sin (4 5^{\circ}) = \frac{2 cos ( x )}{2}

cos (x) sin (4 5^{\circ})

Simplify

sin (4 5^{\circ}) : \frac{2}{2}

sin (4 5^{\circ})

Use the following trivial identity:

sin (4 5^{\circ}) = \frac{2}{2}

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{2}{2}

= \frac{2}{2} cos (x)

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{2 cos ( x )}{2}

= \frac{2 sin ( x )}{2} + \frac{2 cos ( x )}{2}

Apply rule

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{2 sin ( x ) + 2 cos ( x )}{2}

= \frac{2 sin ( x ) + 2 cos ( x )}{2}

= 2 \frac{2 sin ( x ) + 2 cos ( x )}{2}

Simplify

2 \frac{2 sin ( x ) + 2 cos ( x )}{2} : sin (x) + cos (x)

2 \frac{2 sin ( x ) + 2 cos ( x )}{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{( 2 sin ( x ) + 2 cos ( x ) ) 2}{2}

Cancel

\frac{( 2 sin ( x ) + 2 cos ( x ) ) 2}{2} : \frac{2 sin ( x ) + 2 cos ( x )}{2}

\frac{( 2 sin ( x ) + 2 cos ( x ) ) 2}{2}

Apply radical rule:

2 = 2^{\frac{1}{2}}

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

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{2 ^{\frac{1}{2}}}{2 ^{1}} = \frac{1}{2 ^{1 - \frac{1}{2}}}

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

Subtract the numbers:

1 - \frac{1}{2} = \frac{1}{2}

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

Apply radical rule:

2^{\frac{1}{2}} = 2

= \frac{2 sin ( x ) + 2 cos ( x )}{2}

= \frac{2 sin ( x ) + 2 cos ( x )}{2}

Factor out common term

2

= \frac{2 ( sin ( x ) + cos ( x ) )}{2}

Cancel the common factor:

2

= sin (x) + cos (x)

= sin (x) + cos (x)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is sqrt(2)sin(x+45)=sin(x)+cos(x) ?
The answer to whether sqrt(2)sin(x+45)=sin(x)+cos(x) is True