geometry and trig ( please help )

[Ojisan]

Please Help : I am so stuck

Hi, could someone please help me with the following , i have been trying to
do these and am stuck..

the questions are as follows

a) Calculate the exact value of the following ( calculators may not be used )

cos ec² 30˚-sec² 45˚
tan² 60˚

b) Make use of a calculator to determine the following:

20(cot 60˚30'-cosec 2010˚)

c) Make use of a trigonometric identities to prove the following:

sec² θ.cot² θ -1 = cot² θ.

please help

 

[Novice]

Hi, could someone please help me with the following , i have been trying to
do these and am stuck..

the questions are as follows

a) Calculate the exact value of the following ( calculators may not be used )

cos ec² 30˚-sec² 45˚
tan² 60˚

b) Make use of a calculator to determine the following:

20(cot 60˚30'-cosec 2010˚)

c) Make use of a trigonometric identities to prove the following:

sec² θ.cot² θ -1 = cot² θ.

please help

Boy there is nothing to do. I think you hardly know your subject to ask such questions. The first question has T-ratios of standard angles which you should know ( if not, get to know them), and therefore calculators are not allowed. For the second question, since the angles are not standard, you do need a calculator. As for the third, ever head of Trigonometric identities? One of the three basic identities s used here. Find out which.

 

[HallsofIvy]

Hi, could someone please help me with the following , i have been trying to
do these and am stuck..

the questions are as follows

a) Calculate the exact value of the following ( calculators may not be used )

cos ec² 30˚-sec² 45˚
tan² 60˚

I can't make heads or tails of this. Is this supposed to be one fraction? What is the "\(\displaystyle ec^2\)" supposed to mean?

In any case, 30 and 60 degrees are easy because if you take an equilateral triangle (you can take all sides to be of length 1) and drop a perpendicular from one angle to the opposite side, you have two right triangles with one leg, opposite the 60 degree angle, of length 1, the other, opposite the 30 degree angle, of length 1/2. You can use the Pythagorean theorem to find the length of the hypotenuse and so the trig ratios.

b) Make use of a calculator to determine the following:

20(cot 60˚30'-cosec 2010˚)

Okay, have you done that? If not, what is the difficulty?

c) Make use of a trigonometric identities to prove the following:

sec² θ.cot² θ -1 = cot² θ.

please help

The first thing I would do is convert everything to sine and cosine, just because I find their identities easier to remember: sec= 1/cos, cot= cos/sin so this is
\(\displaystyle \frac{1}{cos^2\theta}\frac{cos^2\theta}{sin^2\theta}- 1= \frac{cos^2\theta}{sin^2\theta}\)

 

[Ojisan]

I can't make heads or tails of this. Is this supposed to be one fraction? What is the "\(\displaystyle ec^2\)" supposed to mean?

In any case, 30 and 60 degrees are easy because if you take an equilateral triangle (you can take all sides to be of length 1) and drop a perpendicular from one angle to the opposite side, you have two right triangles with one leg, opposite the 60 degree angle, of length 1, the other, opposite the 30 degree angle, of length 1/2. You can use the Pythagorean theorem to find the length of the hypotenuse and so the trig ratios.


Okay, have you done that? If not, what is the difficulty?


The first thing I would do is convert everything to sine and cosine, just because I find their identities easier to remember: sec= 1/cos, cot= cos/sin so this is
\(\displaystyle \frac{1}{cos^2\theta}\frac{cos^2\theta}{sin^2\theta}- 1= \frac{cos^2\theta}{sin^2\theta}\)

----------------------------------

Hi thanks , for helping , i have check and made some mistakes - please see below i fixed it and all are different sums


question (a) calculate the exact value of the following

cosec^2 30˚-sec^2 45˚
tan^2 60˚

question (b) make use of calculator

20(cot 60˚30'-cosec 210˚)

question (c) make use of trionometric identities to prove the following:

sec^2 θ.cot^2 θ-1=cot^2 θ.

Please i have no idea how to do this if someone can just help i have answers but no idea how to
get to them. here are the answers that i got

(a) 2/3
(b) -11.779
(c) LHS=RHS

also i have no idea what cosec^2 is or cosec

really grateful for the help.

 

[HallsofIvy]

I'm puzzled by this. You seem to be saying that you are not taking and have never taken trigonometry but these are all trigonometry problems.

"cosec" (which is short for "cosecant") is "1 over sine" or, in a right triangle, that is "hypotuse over opposite side". And the "cosec^2" means that squared.

My point before, about the equilateral triangle, is that an equilateral triangle has angles of 60 degrees (because it is equilateral, it has all angle the same and 60+ 60+ 60= 180). Drawing a line from one vertex perpendicular to the opposite side divides the equilateral triangle into two right triangles while dividing the angle into two equal (so 30 degrees) angles and divides the opposite side into two equal parts. That is, if we take the lengths of the sides of the equilateral triangle to be "2", we have a right triangle with angles of 30 and 60 degrees with the side opposite the right angle (the hypotenuse) of length 2 and the side opposite the 30 degree angle of length 1. The Pythagorean theorem tells us that the length of the other side, opposite the 60 degree angle, is \(\displaystyle \sqrt{2^2- 1^2}= \sqrt{3}\). That is, sin(30)= 1/2, cos(30)= \(\displaystyle \sqrt{3}/2\), cosec(30)= 2, \(\displaystyle sec(30)= 2/\sqrt{3}= 2\sqrt{3}/3\), \(\displaystyle tan(30)= 1/\sqrt{3}= \sqrt{3}/3\), and \(\displaystyle cotan(30)= \sqrt{3}\). Switching to the other, 60 degree angle, just swaps "near" and "opposite sides" so we swap each function with it "co" function. sin(60)= \(\displaystyle \sqrt{3}/2\), cos(60)= 1/2, sec(60)= 2, \(\displaystyle cos(60)= 2/sqrt{3}= 2\sqrt{3}/3\), cot(60)= 1/\sqrt{3}= \sqrt{3}/3[/tex], and \(\displaystyle tan(60)= \sqrt{3}\).

A 45 degree angle is also easy because 2(45)= 90 so if one acute angle of a right triangle is 45 so is the other and the triangle is "isosceles". That is, if we take one leg to have length 1, so does the other and, again by the Pythagorean theorem, the hypotenuse has length \(\displaystyle \sqrt{1^2+ 1^2}= \sqrt{2}\). Then sin(45)= cos(45)= \(\displaystyle 1/\sqrt{2}= \sqrt{2}/2\), cosec(45)= sec(45)= \(\displaystyle \sqrt{2}\), and tan(45)= cot(45)= 1.

As far as using a calculator is concerned, the difficulty may be that your calculator only has "sine", "cosine", and "tanget" keys. In that case, you need to know that "cosecant= 1/sine", "secant= 1/cosine", and "cotangent= 1/tangent". In other words, you need to use the "sine", "cosine", and "tangent" keys together with the "1/x" key.

 

[sl8on]

For Part A
Do you know how to find the exact values of trig functions from the 30-60-90 and 45-45-90 triangles?
On a 30-60-90 the side opposite 30 is 1; opposite 60 is sqrt(3); and opposite 90 is 2.
On a 45-45-90 each leg (the sides opposite the 45) is 1; the side opposite 90 is sqrt(2).

Now use the definitions for cosecant = hypotenuse/opposite; secant = hypotenuse/adjacent; and tangent = opposite/adjacent.

You are lucky because each of your ratios have 1 in the denominator. Pulling the ratios from the triangles and plugging in you get:
(2^2 - sqrt(2)^2)/(sqrt(3)^2)
Square each and simplify:
(4-2)/3 = 2/3

For Part B
I tried to work it on my calculator and I couldn't get your answer.

For Part C
Start with the left hand side and use the basic trig identities to rewrite it
sec = 1/cos; cot = cos/sin
So we get
1/cos^2 * cos^2/sin^2 - 1
The cosine squares cancel leaving
1/sin^2 - 1
Rewrite 1/sin^2:
csc^2 - 1
Use the Pythagorean Identity 1 + cot^2 = sec^2 and subtract 1 from each side, so now you have
cot^2 = cot^2