Is sin(θ)+cos(θ)=sin(θ+45) ?

The answer to whether sin(θ)+cos(θ)=sin(θ+45) is False

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prove sin(θ)+cos(θ)=sin(θ+45)

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

θ = 0

sin (0) + cos (0) = 1

sin (0) + cos (0)

Use the following trivial identity:

sin (0) = 0

sin (0)

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}

= 0

Use the following trivial identity:

cos (0) = 1

cos (0)

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}

= 1

= 0 + 1

Simplify

= 1

sin (0 + 4 5^{\circ}) = \frac{2}{2} (Decimal : 0.70710 \dots)

sin (0 + 4 5^{\circ})

Simplify

= sin (4 5^{\circ})

Use the following trivial identity:

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

sin (4 5^{\circ})

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}

The two sides are not equal

\Rightarrow False

\frac{sin ( 2 θ )}{sin ( θ )} = 2 cos (θ)

81 - 9 (3 sin (x))^{2} = 1

\frac{1 + sec ( x )}{csc ( x )} = sin (x) + tan (x)

\frac{tan ^{2} ( t )}{sec ( t )} = sec (t) - cot (t)

81 - 9 (3 sin (x))^{2} = 9 cos (x)

Is sin(θ)+cos(θ)=sin(θ+45) ?
The answer to whether sin(θ)+cos(θ)=sin(θ+45) is False