Trigo basic understanding

StudentQns SG

Using the ASTC thing i can attain the values using the general and basic angle of the different TOA CAH SOH stuff and i wanna check with all of you guys whether my concept is firm and solid or weak and flimsy.

take a as thetha

when u take sin(180-a) it is equal to sin(a) for any values of a. however, you have to take note about which segment it is in. for example sin(180-100)=sin(80) also it is equal to sin(100) as 100 will be the general angle while 80 is the basic angle hence they will be equal since sin(basic angle)=sin(general angle) (must take into account for the ASTC here)

thus i can derive that whenever i do something like sin(360-110)=sin(250)=-sin(70)

so it means that the sin(general angle) will give me the ASTC quadrant while when i take the basic angle i wont?

thanks for the help

wee_ws SG

Hi,

sin (250 deg) = sin (180 deg + 70 deg) = -sin (70 deg),

since 250 deg is in the 3rd quadrant in which only tangent of an angle is positive (while others are negative using sine and cosine functions).

In any case, we identify two things:

1. the acute angle (by rewriting, like what we did above),

2. the quadrant of the given angle (which gives us the sign).

Thanks.

Cheers,
Wen Shih

StudentQns SG

thanks wen shih :) havent seen u in a while

StudentQns SG

wait so means that 250 is like the general angle and 70 is the basic angle?

doraemon1234 SG

I always remember it as all (sin cos tan) positive in 1st quadrant, sin positive in 2nd quadrant, tan positive in 3rd quad, cos positive in 4th quadrant. No real need to memorize it imo cuz i can easily check with my calculator by inputing the angles.

wee_ws SG

Dear StudentQns,

Yes, thanks.

Cheers,
Wen Shih

StudentQns SG

yup the ASTC thing

1) also it gets complicated when i hit the cos(2A) and stuff cos after practice it seems that whenever i get a cos(1/2A) or sin(1/2A) i would have to use cos(A)=2cos^2(1/2A)-1 to solve or cos(A)=1-2sin^2(1/2A) to solve.

so its like fractions you will usually use the cosine formulas unless its tangent then its pretty much only one answer you have to use the 2tan(A)/1-tan^2A.

before all that part is relatively easy but once i hit that part i had some problems with answering questions.

2) for example we know that x=3-square root 16. will there be 2 answers because like square root 16 could be -4 or 4? or is there only one answer. im quite confused with this, if possible could you explain this for me? thanks so much

wee_ws SG

Hi,

I will address your 2nd doubt.

If x = 3 - sqrt(16), it is just

x = 3 - 4 = -1.

If we are solving

(x - 3)^2 - 16 = 0, we will have

x - 3 = +/-sqrt(16)

=> x = 3 + sqrt(16) or 3 - sqrt(16)

=> x = 7 or -1.

In general, if x^2 = a (a > 0), then

x = +/- sqrt(a).

Thanks.

Cheers,
Wen Shih

wee_ws SG

Hi,

It will be useful to look at a worked example to clarify your 1st doubt.

It is given that tan A = 4/3 where A is an acute angle.

Find, without using a calculator, the values of

(i) sin 2A,

(ii) cos 2A,

(iii) tan (A/2).

First, sin A = 4/5 and cos A = 3/5, by considering a right-triangle with sides 3, 4, 5 units.

(i) sin 2A = 2 sin A cos A = 2(4/5)(3/5) = 24/25.

(ii) The value of cos 2A can be found with any of the three identities.

cos 2A = (cos A)^2 - (sin A)^2 = (9/25) - (16/25) = -7/25.

Alternatively, cos 2A = 2(cos A)^2 - 1 = 2(9/25) - 1 = -7/25.

(iii) tan A = {2 tan (A/2)} / [1 - {tan (A/2)}^2]

=> {2 tan (A/2)} / [1 - {tan (A/2)}^2] = 4/3

=> 4 - 4 {tan (A/2)}^2 = 6 tan (A/2)

=> 2 {tan (A/2)}^2 + 3 tan (A/2) - 2 = 0

=> {tan (A/2) + 2} {2 tan (A/2) - 1} = 0

=> tan (A/2) = 1/2, since tan (A/2) must be positive.

Thanks.

Cheers,
Wen Shih

wee_ws SG

Hi,

The identities sin A +/- sin B and cos A +/- cos B may be needed to solve problems, like one question that appeared in this year's paper:

find, without using a calculator, the value of cos (15 deg) - cos (75 deg).

Thanks.

Cheers,
Wen Shih