# Evaluate the product of sines: sin1sin2sin3…sin89

• MHB
• lfdahl
In summary: Happy)In summary, evaluating the product $P = \sin 1\sin 2\sin 3...\sin 89$ without the use of a calculator can be solved using the formula $\sin\frac{\pi}{m}\sin\frac{2\pi}{m}\sin\frac{3\pi}{m}...\sin\frac{(m-1)\pi}{m}=\frac{m}{2^{m-1}}$. This was originally posed as a challenge question and was solved using complex roots of unity. By substituting $m = 180$, the product can be simplified to $\sin 1^\circ \sin 2^\circ \sin 3^\circ...\sin 179 lfdahl Gold Member MHB Evaluate without the use of a calculator the product:$P = \sin 1\sin 2\sin 3…\sin 89$(all angles in degrees) Could you give a hint please? (Angel) Theia said: Could you give a hint please? (Angel) Hint: Use the identity:$\sin (x)\sin (60+x)\sin (60-x) = \frac{1}{4}\sin 3x$(Nod) lfdahl said: Evaluate without the use of a calculator the product:$P = \sin 1\sin 2\sin 3…\sin 89$(all angles in degrees) Using the identities $$\sin x\sin(60-x)\sin(60+x)=\frac 14\sin 3x \tag 1$$ $$\sin x\sin(90-x)=\frac 12\sin 2x \tag 2$$ $$\sin 36 = \sqrt{\frac 58 - \frac{\sqrt 5}8} \tag 3$$ $$\sin 72 = \sqrt{\frac 58 + \frac{\sqrt 5}8} \tag 4$$ we get: $$\overbrace{\sin 1\sin2 ...\sin 89}^{89\text{ factors}} =\frac 1 {4^{29}} \overbrace{\sin 3\sin 6...\sin 87}^{29}\cdot \sin 30 \cdot\sin 60 \\ =\frac 1 {4^{29+9}} \overbrace{\sin 9\sin 18...\sin 81}^{9}\cdot \sin^2 30 \cdot\sin^2 60 \\ =\frac 1 {4^{38}} \frac 1{2^4}\overbrace{\sin 18\sin 36\sin 54\sin 72}^{4}\cdot \sin^2 30 \cdot\sin 45 \cdot \sin^2 60 \\ =\frac 1 {4^{38}} \frac 1{2^{4+2}}\overbrace{\sin 36\sin 72}^{2}\cdot \sin^2 30 \cdot\sin 45 \cdot \sin^2 60 \\ =\frac 1{2^{82}}\cdot\sqrt{\frac 58 - \frac{\sqrt 5}8}\sqrt{\frac 58 + \frac{\sqrt 5}8}\cdot \frac 14 \cdot\sqrt{\frac 12} \cdot \frac 34 \\ =\frac 3{2^{86}}\cdot \sqrt{\frac{5}{2^4}}\cdot\sqrt{\frac 12}\\ =\frac {3}{2^{88}}\sqrt{\frac 52}$$ lfdahl said: Evaluate without the use of a calculator the product:$P = \sin 1\sin 2\sin 3…\sin 89$(all angles in degrees) [sp]Another way to do this is to use the formula$\sin\frac{\pi}{m}\sin\frac{2\pi}{m}\sin\frac{3\pi}{m}\cdots\,\sin\frac{(m-1)\pi}{m}=\frac{m}{2^{m-1}}$(posed as a challenge question by greg1313 in https://mathhelpboards.com/challenge-questions-puzzles-28/trigonometric-product-challenge-21867.html, and neatly solved by kaliprasad using complex roots of unity). If you put$m= 180$in that formula then it becomes $$\sin 1^\circ \sin 2^\circ \sin 3^\circ \cdots \sin 179^\circ = \frac{180}{2^{179}}.$$ But$\sin \theta = \sin(180^\circ - \theta)$, and$\sin 90^\circ = 1$, so we can write that as $$\bigl(\sin 1^\circ \sin 2^\circ \sin 3^\circ \cdots \sin 89^\circ\bigr)^2= \frac{180}{2^{179}} = \frac{36\times 5}{2^{179}}.$$ Now take square roots to get $$\sin 1^\circ \sin 2^\circ \sin 3^\circ \cdots \sin 89^\circ = \frac{3}{2^{88}}\sqrt{\frac 52}.$$ [/sp] I like Serena said: Using the identities $$\sin x\sin(60-x)\sin(60+x)=\frac 14\sin 3x \tag 1$$ $$\sin x\sin(90-x)=\frac 12\sin 2x \tag 2$$ $$\sin 36 = \sqrt{\frac 58 - \frac{\sqrt 5}8} \tag 3$$ $$\sin 72 = \sqrt{\frac 58 + \frac{\sqrt 5}8} \tag 4$$ we get: $$\overbrace{\sin 1\sin2 ...\sin 89}^{89\text{ factors}} =\frac 1 {4^{29}} \overbrace{\sin 3\sin 6...\sin 87}^{29}\cdot \sin 30 \cdot\sin 60 \\ =\frac 1 {4^{29+9}} \overbrace{\sin 9\sin 18...\sin 81}^{9}\cdot \sin^2 30 \cdot\sin^2 60 \\ =\frac 1 {4^{38}} \frac 1{2^4}\overbrace{\sin 18\sin 36\sin 54\sin 72}^{4}\cdot \sin^2 30 \cdot\sin 45 \cdot \sin^2 60 \\ =\frac 1 {4^{38}} \frac 1{2^{4+2}}\overbrace{\sin 36\sin 72}^{2}\cdot \sin^2 30 \cdot\sin 45 \cdot \sin^2 60 \\ =\frac 1{2^{82}}\cdot\sqrt{\frac 58 - \frac{\sqrt 5}8}\sqrt{\frac 58 + \frac{\sqrt 5}8}\cdot \frac 14 \cdot\sqrt{\frac 12} \cdot \frac 34 \\ =\frac 3{2^{86}}\cdot \sqrt{\frac{5}{2^4}}\cdot\sqrt{\frac 12}\\ =\frac {3}{2^{88}}\sqrt{\frac 52}$$ Great job, I like Serena! Thankyou very much for a nice solution!(Happy) - - - Updated - - - Opalg said: [sp]Another way to do this is to use the formula$\sin\frac{\pi}{m}\sin\frac{2\pi}{m}\sin\frac{3\pi}{m}\cdots\,\sin\frac{(m-1)\pi}{m}=\frac{m}{2^{m-1}}$(posed as a challenge question by greg1313 in https://mathhelpboards.com/challenge-questions-puzzles-28/trigonometric-product-challenge-21867.html, and neatly solved by kaliprasad using complex roots of unity). If you put$m= 180$in that formula then it becomes $$\sin 1^\circ \sin 2^\circ \sin 3^\circ \cdots \sin 179^\circ = \frac{180}{2^{179}}.$$ But$\sin \theta = \sin(180^\circ - \theta)$, and$\sin 90^\circ = 1\$, so we can write that as $$\bigl(\sin 1^\circ \sin 2^\circ \sin 3^\circ \cdots \sin 89^\circ\bigr)^2= \frac{180}{2^{179}} = \frac{36\times 5}{2^{179}}.$$ Now take square roots to get $$\sin 1^\circ \sin 2^\circ \sin 3^\circ \cdots \sin 89^\circ = \frac{3}{2^{88}}\sqrt{\frac 52}.$$

[/sp]

Thankyou very much, Opalg for another exemplary solution!

## 1. What is the product of sines for the values 1 through 89?

The product of sines for the values 1 through 89 is approximately 0.9999999999999997.

## 2. How is the product of sines calculated?

The product of sines is calculated by multiplying the sine values of each angle from 1 to 89 degrees.

## 3. Why is the product of sines important in mathematics?

The product of sines is important in mathematics because it is used to solve various problems in trigonometry, such as finding the values of complex trigonometric equations.

## 4. Can the product of sines be greater than 1?

No, the product of sines cannot be greater than 1. This is because the sine function is bounded between -1 and 1, and when multiplied together, the product will also fall within this range.

## 5. What is the geometric interpretation of the product of sines?

The geometric interpretation of the product of sines is the area of the inscribed polygon in a unit circle. Each angle from 1 to 89 degrees represents a side of the polygon, and the product of sines represents the area enclosed by these sides.

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