Studdy AI Logo
Math

QuestionProve that sin⁡60∘⋅tan⁡45∘⋅tan⁡30∘−sin⁡240∘⋅tan⁡315∘⋅tan⁡210∘=0\sin 60^{\circ} \cdot \tan 45^{\circ} \cdot \tan 30^{\circ} - \sin 240^{\circ} \cdot \tan 315^{\circ} \cdot \tan 210^{\circ} = 0.

Studdy Solution

STEP 1

Assumptions1. We are working in degrees, not radians. . We know the values of sine and tangent for the specific angles given in the problem (30°,45°,60°,210°,240°, and315°).
3. We know that sin⁡240∘=−sin⁡60∘\sin240^{\circ} = -\sin60^{\circ} and tan⁡210∘=−tan⁡30∘\tan210^{\circ} = -\tan30^{\circ} and tan⁡315∘=−tan⁡45∘\tan315^{\circ} = -\tan45^{\circ} due to the periodicity and symmetry of the sine and tangent functions.

STEP 2

First, we need to substitute the known values of sine and tangent for the given angles.
sin⁡60∘=2\sin60^{\circ} = \frac{\sqrt{}}{2}tan⁡45∘=1\tan45^{\circ} =1tan⁡30∘=1\tan30^{\circ} = \frac{1}{\sqrt{}}sin⁡240∘=−sin⁡60∘=−2\sin240^{\circ} = -\sin60^{\circ} = -\frac{\sqrt{}}{2}tan⁡315∘=−tan⁡45∘=−1\tan315^{\circ} = -\tan45^{\circ} = -1tan⁡210∘=−tan⁡30∘=−1\tan210^{\circ} = -\tan30^{\circ} = -\frac{1}{\sqrt{}}

STEP 3

Now, substitute these values into the original equation.
32⋅1⋅13−(−32)⋅(−1)⋅(−13)\frac{\sqrt{3}}{2} \cdot1 \cdot \frac{1}{\sqrt{3}} - \left(-\frac{\sqrt{3}}{2}\right) \cdot \left(-1\right) \cdot \left(-\frac{1}{\sqrt{3}}\right)

STEP 4

implify the equation.
12−12\frac{1}{2} - \frac{1}{2}

STEP 5

Calculate the result.
12−12=0\frac{1}{2} - \frac{1}{2} =0Hence, we have proved thatsin⁡60∘⋅tan⁡45∘⋅tan⁡30∘−sin⁡240∘⋅tan⁡315∘⋅tan⁡210∘=0\sin60^{\circ} \cdot \tan45^{\circ} \cdot \tan30^{\circ}-\sin240^{\circ} \cdot \tan315^{\circ} \cdot \tan210^{\circ}=0

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord