Math  /  Trigonometry

Question4. Determine the exact value of each trigonometric expression. a) sin30×tan60cos30\sin 30^{\circ} \times \tan 60^{\circ}-\cos 30^{\circ} c) tan230cos245\tan ^{2} 30^{\circ}-\cos ^{2} 45^{\circ} b) 2cos45×sin452 \cos 45^{\circ} \times \sin 45^{\circ} d) 1sin45cos451-\frac{\sin 45^{\circ}}{\cos 45^{\circ}}
5. Using exact values, show that sin2θ+cos2θ=1\sin ^{2} \theta+\cos ^{2} \theta=1 for each angle. a) θ=30\theta=30^{\circ} b) θ=45\theta=45^{\circ} c) θ=60\theta=60^{\circ}
6. Using exact values, show that sinθcosθ=tanθ\frac{\sin \theta}{\cos \theta}=\tan \theta for each angle. a) θ=30\theta=30^{\circ} b) θ=45\theta=45^{\circ} c) θ=60\theta=60^{\circ}

Studdy Solution

STEP 1

What is this asking? We're going to calculate some trigonometric expressions using special angle values and then verify two super important trigonometric identities! Watch out! Remember those special angle values!
Also, don't mix up sin \sin and cos \cos , and make sure you're squaring the *entire* trig function, not just the angle.

STEP 2

1. Evaluate Trigonometric Expressions
2. Verify Identity 1
3. Verify Identity 2

STEP 3

Alright, let's **start** with 4a: sin30tan60cos30\sin 30^{\circ} \cdot \tan 60^{\circ} - \cos 30^{\circ}.
Remember our special angles?
We have sin30=12 \sin 30^{\circ} = \frac{1}{2} , tan60=3 \tan 60^{\circ} = \sqrt{3} , and cos30=32 \cos 30^{\circ} = \frac{\sqrt{3}}{2} .

STEP 4

Let's **plug** those in: 12332 \frac{1}{2} \cdot \sqrt{3} - \frac{\sqrt{3}}{2} .
This simplifies to 3232=0 \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} = 0 .
Boom!

STEP 5

Next up, 4b: 2cos45sin45 2 \cos 45^{\circ} \cdot \sin 45^{\circ} .
We know cos45=22 \cos 45^{\circ} = \frac{\sqrt{2}}{2} and sin45=22 \sin 45^{\circ} = \frac{\sqrt{2}}{2} .

STEP 6

**Substituting**, we get 22222=224=1 2 \cdot \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = 2 \cdot \frac{2}{4} = 1 .
Nice!

STEP 7

Now for 4c: tan230cos245 \tan^2 30^{\circ} - \cos^2 45^{\circ} .
We have tan30=13 \tan 30^{\circ} = \frac{1}{\sqrt{3}} and cos45=22 \cos 45^{\circ} = \frac{\sqrt{2}}{2} .

STEP 8

**Plugging in** gives us (13)2(22)2=1324=1312=2636=16 (\frac{1}{\sqrt{3}})^2 - (\frac{\sqrt{2}}{2})^2 = \frac{1}{3} - \frac{2}{4} = \frac{1}{3} - \frac{1}{2} = \frac{2}{6} - \frac{3}{6} = -\frac{1}{6} .
Perfect!

STEP 9

Finally, 4d: 1sin45cos45 1 - \frac{\sin 45^{\circ}}{\cos 45^{\circ}} .
We know sin45=22 \sin 45^{\circ} = \frac{\sqrt{2}}{2} and cos45=22 \cos 45^{\circ} = \frac{\sqrt{2}}{2} .

STEP 10

**Substituting**, we get 12222=11=0 1 - \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1 - 1 = 0 .
Awesome!

STEP 11

Let's **prove** sin2θ+cos2θ=1\sin^2 \theta + \cos^2 \theta = 1.
For θ=30 \theta = 30^{\circ} , we have (12)2+(32)2=14+34=1 (\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2 = \frac{1}{4} + \frac{3}{4} = 1 .
It works!

STEP 12

For θ=45 \theta = 45^{\circ} , we have (22)2+(22)2=24+24=1 (\frac{\sqrt{2}}{2})^2 + (\frac{\sqrt{2}}{2})^2 = \frac{2}{4} + \frac{2}{4} = 1 .
Fantastic!

STEP 13

For θ=60 \theta = 60^{\circ} , we have (32)2+(12)2=34+14=1 (\frac{\sqrt{3}}{2})^2 + (\frac{1}{2})^2 = \frac{3}{4} + \frac{1}{4} = 1 .
It's true!

STEP 14

Let's **show** sinθcosθ=tanθ \frac{\sin \theta}{\cos \theta} = \tan \theta .
For θ=30 \theta = 30^{\circ} , sin30cos30=1232=13=tan30 \frac{\sin 30^{\circ}}{\cos 30^{\circ}} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \tan 30^{\circ} .
Yes!

STEP 15

For θ=45 \theta = 45^{\circ} , sin45cos45=2222=1=tan45 \frac{\sin 45^{\circ}}{\cos 45^{\circ}} = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1 = \tan 45^{\circ} .
Amazing!

STEP 16

For θ=60 \theta = 60^{\circ} , sin60cos60=3212=3=tan60 \frac{\sin 60^{\circ}}{\cos 60^{\circ}} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3} = \tan 60^{\circ} .
We did it!

STEP 17

4a: 0, 4b: 1, 4c: 16-\frac{1}{6}, 4d: 0.
We also verified that sin2θ+cos2θ=1\sin^2 \theta + \cos^2 \theta = 1 and sinθcosθ=tanθ\frac{\sin \theta}{\cos \theta} = \tan \theta for θ=30 \theta = 30^{\circ} , 45 45^{\circ} , and 60 60^{\circ} .

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