CEBS95 Group Title tan^2(60) + 4cos^2(45) - 9tan^2(30) - 8sin^2(30) The answer is zero but why!? i need step by step solution please one year ago one year ago

why can't you just plug the whole thing into your calculator?

2. CEBS95 Group Title

because in the exam they already give you the answer, you have to do the procedure

3. Callisto Group Title

tan60 = $$\sqrt{3}$$ cos45 = $$\frac{1}{\sqrt2}$$ tan30 = $$\frac{1}{\sqrt3}$$ sin30 = 1/2 These are all special angles and you should remember them.

4. CEBS95 Group Title

yes i know them but i just continue to get 6 instead of 0

I'm getting zero just fine though. note the squares.

6. CEBS95 Group Title

tan^2(60) = 3 ? 4cos^2(45) = 4 ?

7. Callisto Group Title

4cos^2(45) = 4 (1/2) = ...?

8. CEBS95 Group Title

my math base is not good, that's the reason i wrote step by step solution

9. Callisto Group Title

$cos45 =\frac{1}{\sqrt2}$$cos^245 =(\frac{1}{\sqrt2})^2 = \frac{1}{2}$$4cos45 =4(\frac{1}{2}) = ...?$

10. CEBS95 Group Title

4 (1/2) = 2

11. Callisto Group Title

Yes.

12. winterfez Group Title

$\frac{ \sin ^{2} (60)}{ \cos ^{2}(60)}+4\cos ^{2}(45)-9\frac{ \sin ^{2}(30) }{ \cos ^{2}(30) }-8\sin ^{2}(30)$

13. CEBS95 Group Title

so, tan^2(60) = 3

14. Callisto Group Title

tan^2(60) = 3 <- Yes

15. winterfez Group Title

|dw:1355048071288:dw|

16. CEBS95 Group Title

i found in my book that cos 45 = $\frac{ \sqrt{2} }{ 2}$

17. Callisto Group Title

$\frac{1}{\sqrt2}=\frac{1}{\sqrt2}\times \frac{\sqrt2}{\sqrt2}=\frac{\sqrt2}{\sqrt2^2} = \frac{\sqrt2}{2}$ Just rationalization.

18. CEBS95 Group Title

9tan^2(30) = $\tan( 30) = \frac{ \sqrt{3} }{ 3 } ; \tan ^{2}(30) = 3/9 = 1/3$

19. CEBS95 Group Title

times 9 = 3 right?

20. Callisto Group Title

Yes.

21. CEBS95 Group Title

8 sin ^2 (30) = 4 ?

22. CEBS95 Group Title

no, i think it 8

23. CEBS95 Group Title

9tan^2(30) - 8sin^2(30) = 3 - 8 = -5, at the end that will give me 5- (-5) that equals 10

24. Callisto Group Title

8 sin ^2 (30) = 8 (1/2)^2 = 8 (1/4) = ...

25. CEBS95 Group Title

2

26. CEBS95 Group Title

still, 3-2 = 1, 5-1 = 4

27. winterfez Group Title

|dw:1355049445375:dw| look at this one to see how i get 0

28. Callisto Group Title

tan^2(60) + 4cos^2(45) - 9tan^2(30) - 8sin^2(30) tan^2(60) = $$\sqrt3^2$$ = ... 4cos^2(45) = 4$$(\frac{1}{\sqrt2})^2$$ = 4(1/2) = ... - 9tan^2(30) = -9 $$(\frac{1}{\sqrt3})^2$$ = -9 (1/3) = ... - 8sin^2(30) = -8 (1/2)^2 = -8 (1/4) = ... Sum up the four answers.

29. winterfez Group Title

|dw:1355049643294:dw|

30. CEBS95 Group Title

zero! nice thanks calisto

31. Callisto Group Title

You're welcome :)

32. Callisto Group Title

| 30 | 45 | 60 ------------------------------------ sin | $$\frac{1}{2}$$ | $$\frac{1}{\sqrt2} /\frac{\sqrt2}{2}$$ | $$\frac{\sqrt3}{2}$$ ------------------------------------ cos | $$\frac{\sqrt3}{2}$$ | $$\frac{1}{\sqrt2} /\frac{\sqrt2}{2}$$ | $$\frac{1}{2}$$ ------------------------------------ tan | $$\frac{1}{\sqrt3} /\frac{\sqrt3}{3}$$ | 1 | $$\sqrt{3}$$

33. Callisto Group Title

Memorize the table..

34. CEBS95 Group Title

yes, with practice i will memorise the terms