4) The cosine of 75° has an exact value of [tex]\frac{\sqrt{6}-\sqrt{2}}{4}[/tex].
5) The tangent of [tex]-\frac{\pi}{12}[/tex] radians has an exact value of [tex]-\frac{\sqrt{3}}{3-\sqrt{3}}[/tex].
6) The proposition is true.
Procedure - Determine of exact values by sines and cosines by trigonometric formulas
4) Exact value of the cosine of 75°
We can determine this cosine by means of the following trigonometric formula:
[tex]\cos (\alpha + \beta) = \cos \alpha \cdot \cos \beta - \sin \alpha \cdot \sin \beta[/tex] (1)
Where [tex]\alpha[/tex], [tex]\beta[/tex] are angles expressed in degrees.
Resolution of the trigonometric expression
If we know that [tex]\alpha = 45^{\circ}[/tex], [tex]\beta = 30^{\circ}[/tex], then we have that the cosine of 75° is:
[tex]\cos 75^{\circ} = \cos 45^{\circ}\cdot \cos 30^{\circ} - \sin 45^{\circ}\cdot \sin 30^{\circ}[/tex]
[tex]\cos 75^{\circ} = \left(\frac{\sqrt{2}}{2}\right)\cdot \left(\frac{\sqrt{3}}{2} \right) - \left(\frac{\sqrt{2}}{2} \right)\cdot \left(\frac{1}{2} \right)[/tex]
[tex]\cos 75^{\crc} = \frac{\sqrt{6}}{4} -\frac{\sqrt{2}}{4}[/tex]
[tex]\cos 75^{\circ} = \frac{\sqrt{6}-\sqrt{2}}{4}[/tex]
The cosine of 75° has an exact value of [tex]\frac{\sqrt{6}-\sqrt{2}}{4}[/tex]. [tex]\blacksquare[/tex]
5) Exact value of the tangent of -π/12 radians
We can determine this tangent by means of the following trigonometric formulas:
[tex]\tan (\alpha - \beta) = \frac{\tan \alpha \cdot \tan \beta}{1+\tan \alpha \cdot \tan \beta}[/tex] (2)
[tex]\tan (-\alpha) = -\tan \alpha[/tex] (3)
Where [tex]\alpha[/tex], [tex]\beta[/tex] are angles measured in radians.
Resolution of the trigonometric expression
If we know that [tex]\alpha = -\frac{\pi}{4}[/tex] and [tex]\beta = \frac{\pi}{6}[/tex], then the tangent of [tex]-\frac{\pi}{12}[/tex] radians is:
[tex]\tan\left(-\frac{\pi}{12} \right) = \frac{\tan \left(-\frac{\pi}{4} \right)\cdot \tan \left(\frac{\pi}{6} \right)}{1+\tan \left(-\frac{\pi}{4} \right)\cdot \tan \left(\frac{\pi}{6}\right)}[/tex]
[tex]\tan \left(-\frac{\pi}{12} \right) = \frac{(-1)\cdot \left(\frac{\sqrt{3}}{3} \right)}{1 + (-1)\cdot \left(\frac{\sqrt{3}}{3} \right)}[/tex]
[tex]\tan \left(-\frac{\pi}{12} \right) = -\frac{\sqrt{3}}{3-\sqrt{3}}[/tex]
The tangent of [tex]-\frac{\pi}{12}[/tex] radians has an exact value of [tex]-\frac{\sqrt{3}}{3-\sqrt{3}}[/tex]. [tex]\blacksquare[/tex]
Procedure: Simplification of a trigonometric expression by trigonometric formula
6) Resolution of a trigonometric formula
In this part we are going to simplify by algebraic and trigonometric means an expression, whose procedure is shown below:
- [tex]\left(\frac{1-\tan^{2}x}{\sec^{2}x} \right)[/tex] Given
- [tex]\left(1-\frac{\sin^{2}x}{\cos^{2}x} \right)\cdot \cos^{2}x[/tex] [tex]\sec x = \frac{1}{\cos x}[/tex], [tex]\tan x = \frac{\sin x}{\cos x}[/tex]
- [tex]\cos^{2}x - \sin^{2} x[/tex] Distributive property/Existence of multiplicative inverse
- [tex]\cos 2x[/tex] [tex]\cos 2x = \cos^{2} x - \sin^{2} x[/tex]/Result
The proposition is true. [tex]\blacksquare[/tex]
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