Is there a simple way to calculate $sin frac{3pi}{10}-sin frac{pi}{10}$? [duplicate]
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Proving trigonometric equation $cos(36^circ) - cos(72^circ) = 1/2$ [duplicate]
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I know how to find the exact value of $sin frac{pi}{10}$ using double and triple angle formulas and the fact that $frac{5pi}{10}=frac{pi}{2}$ but it maybe too complicated for high school students. Is there an easier way that I do not see? The answer is $0.5$.
Unfortunately I did not see a clear answer among posted answers to my question but thanks to @labbhattacharjee, the main idea is to multiply and divide by $2 cos 18°$.
Thus $large{sin 54°-sin 18°=frac{2 sin 54°cos 18°-2 sin 18°cos 18°}{2 cos 18°}=frac{sin 72° + sin 36°-sin 36°}{2 cos 18°}=frac{sin 72°}{2cos 18°}=0.5}$
trigonometry
asked Nov 14 at 5:13
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Nov 14 at 8:19
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This question already has an answer here:
Proving trigonometric equation $cos(36^circ) - cos(72^circ) = 1/2$ [duplicate]
2 answers
I know how to find the exact value of $sin frac{pi}{10}$ using double and triple angle formulas and the fact that $frac{5pi}{10}=frac{pi}{2}$ but it maybe too complicated for high school students. Is there an easier way that I do not see? The answer is $0.5$.
Unfortunately I did not see a clear answer among posted answers to my question but thanks to @labbhattacharjee, the main idea is to multiply and divide by $2 cos 18°$.
Thus $large{sin 54°-sin 18°=frac{2 sin 54°cos 18°-2 sin 18°cos 18°}{2 cos 18°}=frac{sin 72° + sin 36°-sin 36°}{2 cos 18°}=frac{sin 72°}{2cos 18°}=0.5}$
trigonometry
asked Nov 14 at 5:13
Vasya
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marked as duplicate by lab bhattacharjee trigonometry
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Nov 14 at 8:19
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up vote
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down vote
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This question already has an answer here:
Proving trigonometric equation $cos(36^circ) - cos(72^circ) = 1/2$ [duplicate]
2 answers
I know how to find the exact value of $sin frac{pi}{10}$ using double and triple angle formulas and the fact that $frac{5pi}{10}=frac{pi}{2}$ but it maybe too complicated for high school students. Is there an easier way that I do not see? The answer is $0.5$.
Unfortunately I did not see a clear answer among posted answers to my question but thanks to @labbhattacharjee, the main idea is to multiply and divide by $2 cos 18°$.
Thus $large{sin 54°-sin 18°=frac{2 sin 54°cos 18°-2 sin 18°cos 18°}{2 cos 18°}=frac{sin 72° + sin 36°-sin 36°}{2 cos 18°}=frac{sin 72°}{2cos 18°}=0.5}$
trigonometry
asked Nov 14 at 5:13
Vasya
3,2391515
This question already has an answer here:
Proving trigonometric equation $cos(36^circ) - cos(72^circ) = 1/2$ [duplicate]
2 answers
I know how to find the exact value of $sin frac{pi}{10}$ using double and triple angle formulas and the fact that $frac{5pi}{10}=frac{pi}{2}$ but it maybe too complicated for high school students. Is there an easier way that I do not see? The answer is $0.5$.
Unfortunately I did not see a clear answer among posted answers to my question but thanks to @labbhattacharjee, the main idea is to multiply and divide by $2 cos 18°$.
Thus $large{sin 54°-sin 18°=frac{2 sin 54°cos 18°-2 sin 18°cos 18°}{2 cos 18°}=frac{sin 72° + sin 36°-sin 36°}{2 cos 18°}=frac{sin 72°}{2cos 18°}=0.5}$
This question already has an answer here:
Proving trigonometric equation $cos(36^circ) - cos(72^circ) = 1/2$ [duplicate]
2 answers
trigonometry
trigonometry
asked Nov 14 at 5:13
Vasya
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asked Nov 14 at 5:13
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edited Nov 14 at 14:34
asked Nov 14 at 5:13
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asked Nov 14 at 5:13
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asked Nov 14 at 5:13
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Nov 14 at 8:19
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3 Answers
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Observe
begin{align}
sin x-sin y=2sinleft(frac{x-y}{2}right)cosleft(frac{x+y}{2}right)
end{align}
then it follows
begin{align}
sin frac{3pi}{10}-sinfrac{pi}{10} = 2sinfrac{pi}{10}cosfrac{2pi}{10} = 2sinfrac{pi}{10}left(cos^2frac{pi}{10}-sin^2frac{pi}{10} right).
end{align}
answered Nov 14 at 5:20
Jacky Chong
17.1k21027
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begin{align}
sin frac{3pi}{10}&=cos(frac{pi}{2} - frac{3pi}{10})\
&Rightarrow cos(frac{pi}{2} - frac{3pi}{10})-sinfrac{pi}{10} \
&=cosfrac{pi}{5} -sinfrac{pi}{10}
end{align}
answered Nov 14 at 5:33
1ENİGMA1
940316
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That is
$$cosfrac{pi}{5}-cosfrac{2pi}5=-cosfrac{4pi}{5}-cosfrac{2pi}5=A$$
say.
Then
$$Asinfrac{pi}5=-sinfrac{pi}5left(cosfrac{4pi}{5}+cosfrac{2pi}5right)=-frac12left(sinfrac{5pi}5-sinfrac{3pi}5+sinfrac{3pi}5-
sinfrac{pi}5right)=frac12sinfracpi5$$
etc.
answered Nov 14 at 5:55
Lord Shark the Unknown
97k958128
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Observe
begin{align}
sin x-sin y=2sinleft(frac{x-y}{2}right)cosleft(frac{x+y}{2}right)
end{align}
then it follows
begin{align}
sin frac{3pi}{10}-sinfrac{pi}{10} = 2sinfrac{pi}{10}cosfrac{2pi}{10} = 2sinfrac{pi}{10}left(cos^2frac{pi}{10}-sin^2frac{pi}{10} right).
end{align}
answered Nov 14 at 5:20
Jacky Chong
17.1k21027
add a comment |
up vote
2
down vote
Observe
begin{align}
sin x-sin y=2sinleft(frac{x-y}{2}right)cosleft(frac{x+y}{2}right)
end{align}
then it follows
begin{align}
sin frac{3pi}{10}-sinfrac{pi}{10} = 2sinfrac{pi}{10}cosfrac{2pi}{10} = 2sinfrac{pi}{10}left(cos^2frac{pi}{10}-sin^2frac{pi}{10} right).
end{align}
answered Nov 14 at 5:20
Jacky Chong
17.1k21027
add a comment |
up vote
2
down vote
up vote
2
down vote
Observe
begin{align}
sin x-sin y=2sinleft(frac{x-y}{2}right)cosleft(frac{x+y}{2}right)
end{align}
then it follows
begin{align}
sin frac{3pi}{10}-sinfrac{pi}{10} = 2sinfrac{pi}{10}cosfrac{2pi}{10} = 2sinfrac{pi}{10}left(cos^2frac{pi}{10}-sin^2frac{pi}{10} right).
end{align}
answered Nov 14 at 5:20
Jacky Chong
17.1k21027
Observe
begin{align}
sin x-sin y=2sinleft(frac{x-y}{2}right)cosleft(frac{x+y}{2}right)
end{align}
then it follows
begin{align}
sin frac{3pi}{10}-sinfrac{pi}{10} = 2sinfrac{pi}{10}cosfrac{2pi}{10} = 2sinfrac{pi}{10}left(cos^2frac{pi}{10}-sin^2frac{pi}{10} right).
end{align}
answered Nov 14 at 5:20
Jacky Chong
17.1k21027
answered Nov 14 at 5:20
Jacky Chong
17.1k21027
answered Nov 14 at 5:20
Jacky Chong
17.1k21027
answered Nov 14 at 5:20
Jacky Chong
17.1k21027
17.1k21027
add a comment |
add a comment |
up vote
0
down vote
begin{align}
sin frac{3pi}{10}&=cos(frac{pi}{2} - frac{3pi}{10})\
&Rightarrow cos(frac{pi}{2} - frac{3pi}{10})-sinfrac{pi}{10} \
&=cosfrac{pi}{5} -sinfrac{pi}{10}
end{align}
answered Nov 14 at 5:33
1ENİGMA1
940316
add a comment |
up vote
0
down vote
begin{align}
sin frac{3pi}{10}&=cos(frac{pi}{2} - frac{3pi}{10})\
&Rightarrow cos(frac{pi}{2} - frac{3pi}{10})-sinfrac{pi}{10} \
&=cosfrac{pi}{5} -sinfrac{pi}{10}
end{align}
answered Nov 14 at 5:33
1ENİGMA1
940316
add a comment |
up vote
0
down vote
up vote
0
down vote
begin{align}
sin frac{3pi}{10}&=cos(frac{pi}{2} - frac{3pi}{10})\
&Rightarrow cos(frac{pi}{2} - frac{3pi}{10})-sinfrac{pi}{10} \
&=cosfrac{pi}{5} -sinfrac{pi}{10}
end{align}
answered Nov 14 at 5:33
1ENİGMA1
940316
begin{align}
sin frac{3pi}{10}&=cos(frac{pi}{2} - frac{3pi}{10})\
&Rightarrow cos(frac{pi}{2} - frac{3pi}{10})-sinfrac{pi}{10} \
&=cosfrac{pi}{5} -sinfrac{pi}{10}
end{align}
answered Nov 14 at 5:33
1ENİGMA1
940316
answered Nov 14 at 5:33
1ENİGMA1
940316
answered Nov 14 at 5:33
1ENİGMA1
940316
answered Nov 14 at 5:33
1ENİGMA1
940316
940316
add a comment |
add a comment |
up vote
0
down vote
That is
$$cosfrac{pi}{5}-cosfrac{2pi}5=-cosfrac{4pi}{5}-cosfrac{2pi}5=A$$
say.
Then
$$Asinfrac{pi}5=-sinfrac{pi}5left(cosfrac{4pi}{5}+cosfrac{2pi}5right)=-frac12left(sinfrac{5pi}5-sinfrac{3pi}5+sinfrac{3pi}5-
sinfrac{pi}5right)=frac12sinfracpi5$$
etc.
answered Nov 14 at 5:55
Lord Shark the Unknown
97k958128
add a comment |
up vote
0
down vote
That is
$$cosfrac{pi}{5}-cosfrac{2pi}5=-cosfrac{4pi}{5}-cosfrac{2pi}5=A$$
say.
Then
$$Asinfrac{pi}5=-sinfrac{pi}5left(cosfrac{4pi}{5}+cosfrac{2pi}5right)=-frac12left(sinfrac{5pi}5-sinfrac{3pi}5+sinfrac{3pi}5-
sinfrac{pi}5right)=frac12sinfracpi5$$
etc.
answered Nov 14 at 5:55
Lord Shark the Unknown
97k958128
add a comment |
up vote
0
down vote
up vote
0
down vote
That is
$$cosfrac{pi}{5}-cosfrac{2pi}5=-cosfrac{4pi}{5}-cosfrac{2pi}5=A$$
say.
Then
$$Asinfrac{pi}5=-sinfrac{pi}5left(cosfrac{4pi}{5}+cosfrac{2pi}5right)=-frac12left(sinfrac{5pi}5-sinfrac{3pi}5+sinfrac{3pi}5-
sinfrac{pi}5right)=frac12sinfracpi5$$
etc.
answered Nov 14 at 5:55
Lord Shark the Unknown
97k958128
That is
$$cosfrac{pi}{5}-cosfrac{2pi}5=-cosfrac{4pi}{5}-cosfrac{2pi}5=A$$
say.
Then
$$Asinfrac{pi}5=-sinfrac{pi}5left(cosfrac{4pi}{5}+cosfrac{2pi}5right)=-frac12left(sinfrac{5pi}5-sinfrac{3pi}5+sinfrac{3pi}5-
sinfrac{pi}5right)=frac12sinfracpi5$$
etc.
answered Nov 14 at 5:55
Lord Shark the Unknown
97k958128
answered Nov 14 at 5:55
Lord Shark the Unknown
97k958128
answered Nov 14 at 5:55
Lord Shark the Unknown
97k958128
answered Nov 14 at 5:55
Lord Shark the Unknown
97k958128
97k958128
add a comment |
add a comment |
