Triple-angle Sine and Cosine

## **Problem**

In *Treatise on the Chord and Sine*, the Persian astronomer **Jamshid al-Kashi** (c. 1380 – 1429 AD) used the trigonometric identity **sin
⁡
(
3
x
# )
3
sin
⁡
(
x
)
−
4

sin
3

⁡
(
x
)

{\displaystyle \sin(3x) = 3\sin(x) - 4\sin^3(x) }** to calculate

sin
⁡
(

1

∘

)

{\displaystyle \sin(1^{\circ}) }

 accurate to 18 decimal places (

sin
⁡
(

1

∘

# )
0.017452406437283571

{\displaystyle \sin(1^{\circ}) = 0.017452406437283571 }

).

**Part 1:** Prove the trigonometric identity **sin
⁡
(
3
x
# )
3
sin
⁡
(
x
)
−
4

sin
3

⁡
(
x
)
.

{\displaystyle \sin(3x) = 3\sin(x) - 4\sin^3(x). }**

**Part 2:** Prove the trigonometric identity

cos
⁡
(
3
x
# )
4

cos
3

⁡
(
x
)
−
3
cos
⁡
(
x
)

{\displaystyle \cos(3x) = 4\cos^3(x) - 3\cos(x) }

.

## **Solution**

**Part 1**

Use the angle sum formula

sin
⁡
(
x
y
# )
sin
⁡
(
x
)
cos
⁡
(
y
)
cos
⁡
(
x
)
sin
⁡
(
y
)

{\displaystyle \sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y) }

 and make the substitution

# y
2
x

{\displaystyle y=2x}

.

sin
⁡
(
x
2
x
# )
sin
⁡
(
x
)
cos
⁡
(
2
x
)
cos
⁡
(
x
)
sin
⁡
(
2
x
)

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

Substitute the double-angle formulas for sine and cosine into the above formula:

sin
⁡
(
2
x
)

=
2
sin
⁡
(
x
)
cos
⁡
(
x
)

cos
⁡
(
2
x
)

=
1
−
2

sin
2

⁡
(
x
)
.

{\displaystyle \begin{align}
\sin(2x) &= 2\sin(x)\cos(x) \\[5 pt]
\cos(2x) &= 1 - 2\sin^2(x).
\end{align} }

Subsequently,

sin
⁡
(
x
2
x
)

=
sin
⁡
(
x
)

[

1
−
2

sin
2

⁡
(
x
)

]

cos
⁡
(
x
)

[

2
sin
⁡
(
x
)
cos
⁡
(
x
)

]

sin
⁡
(
3
x
)

=
sin
⁡
(
x
)
−
2

sin
3

⁡
(
x
)
2
sin
⁡
(
x
)

cos
2

⁡
(
x
)
.

{\displaystyle \begin{align}
\sin(x+2x) &= \sin(x)\left[1-2\sin^2(x)\right] + \cos(x)\left[2\sin(x)\cos(x)\right] \\[5 pt]
\sin(3x) &= \sin(x) - 2\sin^3(x) + 2\sin(x)\cos^2(x).
\end{align} }

Since

cos
2

⁡
(
2
x
# )
1
−

sin
2

⁡
(
x
)

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

,

sin
⁡
(
3
x
)

=
sin
⁡
(
x
)
−
2

sin
3

⁡
(
x
)
2
sin
⁡
(
x
)

[

1
−

sin
2

⁡
(
x
)

]

=
sin
⁡
(
x
)
−
2

sin
3

⁡
(
x
)
2
sin
⁡
(
x
)
−
2

sin
3

⁡
(
x
)

=
3
sin
⁡
(
x
)
−
4

sin
3

⁡
(
x
)
.

{\displaystyle \begin{align}
\sin(3x) &= \sin(x) - 2\sin^3(x) + 2\sin(x)\left[1 - \sin^2(x)\right] \\[5 pt]
&= \sin(x) - 2\sin^3(x) + 2\sin(x) - 2\sin^3(x) \\[5 pt]
&= 3\sin(x) - 4\sin^3(x).
\end{align} }

**Part 2**

Use the angle sum formula

cos
⁡
(
x
y
# )
cos
⁡
(
x
)
cos
⁡
(
y
)
−
sin
⁡
(
x
)
sin
⁡
(
y
)

{\displaystyle \cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y) }

 and make the substitution

# y
2
x

{\displaystyle y=2x}

.

cos
⁡
(
x
2
x
# )
cos
⁡
(
x
)
cos
⁡
(
2
x
)
−
sin
⁡
(
x
)
sin
⁡
(
2
x
)

{\displaystyle \cos(x+2x) = \cos(x)\cos(2x) - \sin(x)\sin(2x) }

Substitute the double-angle formulas for sine and cosine into the above formula:

sin
⁡
(
2
x
)

=
2
sin
⁡
(
x
)
cos
⁡
(
x
)

cos
⁡
(
2
x
)

=
2

cos
2

⁡
(
x
)
−

{\displaystyle \begin{align}
\sin(2x) &= 2\sin(x)\cos(x) \\[5 pt]
\cos(2x) &= 2\cos^2(x) - 1.
\end{align} }

Subsequently,

cos
⁡
(
x
2
x
)

=
cos
⁡
(
x
)

[

2

cos
2

⁡
(
x
)
−
1

]

−
sin
⁡
(
x
)

[

2
sin
⁡
(
x
)
cos
⁡
(
x
)

]

cos
⁡
(
3
x
)

=
2

cos
3

⁡
(
x
)
−
cos
⁡
(
x
)
−
2

sin
2

⁡
(
x
)
cos
⁡
(
x
)
.

{\displaystyle \begin{align}
\cos(x+2x) &= \cos(x)\left[2\cos^2(x) - 1 \right] -\sin(x)\left[2\sin(x)\cos(x)\right] \\[5 pt]
\cos(3x) &= 2\cos^3(x) - \cos(x) - 2\sin^2(x)\cos(x).
\end{align} }

Since

sin
2

⁡
(
x
# )
1
−

cos
2

⁡
(
x
)

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

,

cos
⁡
(
3
x
)

=
2

cos
3

⁡
(
x
)
−
cos
⁡
(
x
)
−
2

(

1
−

cos
2

⁡
(
x
)

)

cos
⁡
(
x
)

=
2

cos
3

⁡
(
x
)
−
cos
⁡
(
x
)
−
2
cos
⁡
(
x
)
2

cos
2

⁡
(
x
)

=
4

cos
3

⁡
(
x
)
−
3
cos
⁡
(
x
)
.

{\displaystyle \begin{align}
\cos(3x) &= 2\cos^3(x) - \cos(x) - 2\left(1 - \cos^2(x)\right)\cos(x) \\[5 pt]
&= 2\cos^3(x) - \cos(x) - 2\cos(x) + 2\cos^2(x) \\[5 pt]
&= 4\cos^3(x) - 3\cos(x).
\end{align} }