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Sagot :
To show the trigonometric identity:
[tex]\[ \cos \theta \cos \left(60^\circ + \theta \right) \cos \left(60^\circ - \theta \right) = \frac{1}{4} \cos 3\theta, \][/tex]
we can break it down step-by-step:
1. Left-Hand Side:
[tex]\[ \cos \theta \cos \left(60^\circ + \theta\right) \cos \left(60^\circ - \theta\right) \][/tex]
2. Right-Hand Side:
[tex]\[ \frac{1}{4} \cos 3\theta \][/tex]
We need to show that the left-hand side simplifies to the right-hand side.
### Simplifying the Left-Hand Side
First, we recall the cosine addition and subtraction formulas:
[tex]\[ \cos(60^\circ + \theta) = \cos 60^\circ \cos \theta - \sin 60^\circ \sin \theta \][/tex]
[tex]\[ \cos(60^\circ - \theta) = \cos 60^\circ \cos \theta + \sin 60^\circ \sin \theta \][/tex]
Given that:
[tex]\[ \cos 60^\circ = \frac{1}{2} \][/tex]
[tex]\[ \sin 60^\circ = \frac{\sqrt{3}}{2} \][/tex]
We substitute these into our expressions:
[tex]\[ \cos(60^\circ + \theta) = \frac{1}{2} \cos \theta - \frac{\sqrt{3}}{2} \sin \theta \][/tex]
[tex]\[ \cos(60^\circ - \theta) = \frac{1}{2} \cos \theta + \frac{\sqrt{3}}{2} \sin \theta \][/tex]
Next, we substitute these into the left-hand side expression:
[tex]\[ \cos \theta \left( \left( \frac{1}{2} \cos \theta - \frac{\sqrt{3}}{2} \sin \theta \right) \times \left( \frac{1}{2} \cos \theta + \frac{\sqrt{3}}{2} \sin \theta \right) \right) \][/tex]
Now, using the difference of squares identity [tex]\( (a - b)(a + b) = a^2 - b^2 \)[/tex]:
[tex]\[ \left( \frac{1}{2} \cos \theta - \frac{\sqrt{3}}{2} \sin \theta \right) \left( \frac{1}{2} \cos \theta + \frac{\sqrt{3}}{2} \sin \theta \right) = \left( \frac{1}{2} \cos \theta \right)^2 - \left( \frac{\sqrt{3}}{2} \sin \theta \right)^2 \][/tex]
Simplify inside the parenthesis:
[tex]\[ \left( \frac{1}{2} \cos \theta \right)^2 = \frac{1}{4} \cos^2 \theta \][/tex]
[tex]\[ \left( \frac{\sqrt{3}}{2} \sin \theta \right)^2 = \frac{3}{4} \sin^2 \theta \][/tex]
Putting these back into the equation:
[tex]\[ \frac{1}{4} \cos^2 \theta - \frac{3}{4} \sin^2 \theta \][/tex]
Factor out [tex]\( \frac{1}{4} \)[/tex]:
[tex]\[ \frac{1}{4} ( \cos^2 \theta - 3 \sin^2 \theta ) \][/tex]
Now, recall the triple angle formula for cosine:
[tex]\[ \cos 3 \theta = 4 \cos^3 \theta - 3 \cos \theta \][/tex]
For [tex]\( \cos^2 \theta - 3 \sin^2 \theta \)[/tex], we use the Pythagorean identity [tex]\( \sin^2 \theta = 1 - \cos^2 \theta \)[/tex]:
[tex]\[ \cos^2 \theta - 3(1 - \cos^2 \theta) = \cos^2 \theta - 3 + 3 \cos^2 \theta = 4 \cos^2 \theta - 3 \][/tex]
Thus,
[tex]\[ \cos \theta \left( \frac{1}{4} (4 \cos^2 \theta - 3) \right) \][/tex]
Finally,
[tex]\[ \cos \theta \left( \cos 3 \theta \right) = \frac{1}{4} \cos 3 \theta \][/tex]
Thus, we have shown that:
[tex]\[ \cos \theta \cos \left(60^\circ+\theta\right) \cos \left(60^\circ-\theta\right) = \frac{1}{4} \cos 3 \theta \][/tex]
So, the left-hand side indeed simplifies to the right-hand side. This completes the proof of the identity.
[tex]\[ \cos \theta \cos \left(60^\circ + \theta \right) \cos \left(60^\circ - \theta \right) = \frac{1}{4} \cos 3\theta, \][/tex]
we can break it down step-by-step:
1. Left-Hand Side:
[tex]\[ \cos \theta \cos \left(60^\circ + \theta\right) \cos \left(60^\circ - \theta\right) \][/tex]
2. Right-Hand Side:
[tex]\[ \frac{1}{4} \cos 3\theta \][/tex]
We need to show that the left-hand side simplifies to the right-hand side.
### Simplifying the Left-Hand Side
First, we recall the cosine addition and subtraction formulas:
[tex]\[ \cos(60^\circ + \theta) = \cos 60^\circ \cos \theta - \sin 60^\circ \sin \theta \][/tex]
[tex]\[ \cos(60^\circ - \theta) = \cos 60^\circ \cos \theta + \sin 60^\circ \sin \theta \][/tex]
Given that:
[tex]\[ \cos 60^\circ = \frac{1}{2} \][/tex]
[tex]\[ \sin 60^\circ = \frac{\sqrt{3}}{2} \][/tex]
We substitute these into our expressions:
[tex]\[ \cos(60^\circ + \theta) = \frac{1}{2} \cos \theta - \frac{\sqrt{3}}{2} \sin \theta \][/tex]
[tex]\[ \cos(60^\circ - \theta) = \frac{1}{2} \cos \theta + \frac{\sqrt{3}}{2} \sin \theta \][/tex]
Next, we substitute these into the left-hand side expression:
[tex]\[ \cos \theta \left( \left( \frac{1}{2} \cos \theta - \frac{\sqrt{3}}{2} \sin \theta \right) \times \left( \frac{1}{2} \cos \theta + \frac{\sqrt{3}}{2} \sin \theta \right) \right) \][/tex]
Now, using the difference of squares identity [tex]\( (a - b)(a + b) = a^2 - b^2 \)[/tex]:
[tex]\[ \left( \frac{1}{2} \cos \theta - \frac{\sqrt{3}}{2} \sin \theta \right) \left( \frac{1}{2} \cos \theta + \frac{\sqrt{3}}{2} \sin \theta \right) = \left( \frac{1}{2} \cos \theta \right)^2 - \left( \frac{\sqrt{3}}{2} \sin \theta \right)^2 \][/tex]
Simplify inside the parenthesis:
[tex]\[ \left( \frac{1}{2} \cos \theta \right)^2 = \frac{1}{4} \cos^2 \theta \][/tex]
[tex]\[ \left( \frac{\sqrt{3}}{2} \sin \theta \right)^2 = \frac{3}{4} \sin^2 \theta \][/tex]
Putting these back into the equation:
[tex]\[ \frac{1}{4} \cos^2 \theta - \frac{3}{4} \sin^2 \theta \][/tex]
Factor out [tex]\( \frac{1}{4} \)[/tex]:
[tex]\[ \frac{1}{4} ( \cos^2 \theta - 3 \sin^2 \theta ) \][/tex]
Now, recall the triple angle formula for cosine:
[tex]\[ \cos 3 \theta = 4 \cos^3 \theta - 3 \cos \theta \][/tex]
For [tex]\( \cos^2 \theta - 3 \sin^2 \theta \)[/tex], we use the Pythagorean identity [tex]\( \sin^2 \theta = 1 - \cos^2 \theta \)[/tex]:
[tex]\[ \cos^2 \theta - 3(1 - \cos^2 \theta) = \cos^2 \theta - 3 + 3 \cos^2 \theta = 4 \cos^2 \theta - 3 \][/tex]
Thus,
[tex]\[ \cos \theta \left( \frac{1}{4} (4 \cos^2 \theta - 3) \right) \][/tex]
Finally,
[tex]\[ \cos \theta \left( \cos 3 \theta \right) = \frac{1}{4} \cos 3 \theta \][/tex]
Thus, we have shown that:
[tex]\[ \cos \theta \cos \left(60^\circ+\theta\right) \cos \left(60^\circ-\theta\right) = \frac{1}{4} \cos 3 \theta \][/tex]
So, the left-hand side indeed simplifies to the right-hand side. This completes the proof of the identity.
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