Introduction to Algebra using the TI-92 Ó 1997
by Nathan O. Niles
Associate Professor (Retired)
U. S. Naval Academy

15. Inverse Trig Functions

Values of inverse trigonometric functions may be obtained by use of a calculator or from tables. The TI-92 Graphing Calculator has keys SIN-1, COS-1, and TAN-1. They are the ‘second function’ for the SIN, COS, and TAN keys and require the  key to be pressed first. When using a calculator, the ‘mode’ must be radians when a real number or an angle in radians is desired. When an angle in degrees is desired the ‘mode’ must be degrees.

Example. Find the number for each of the following using the TI-92 Graphing Calculator.

(a) Sin-1(0.2337). (b) Sin-1(- 0.567). (c) Cos-1(- 0.9211). (d) Tan-1(1.743).

Solution: Since the answer is to be a number the mode must be radians.

Turn on the TI-92, clear home screen and entry line:   8 

Check the mode:  (if the fourth line is: Angle …….. RADIAN press  )
(if the fourth line is: Angle ………DEGREE press   1  )

(a)  [SIN-1] 0.2337   Displays: .235881

(b)  [SIN-1]  0.567   Displays: - .602859

(c)  [COS-1]  0.9211   Displays: 2.74169

(d)  [TAN-1] 1.743   Displays: 1.04992

Problem. Find the number for each of the following using the TI-92 Graphing Calculator.

(a) Sin-1(0.7833). (b) Sin-1(- 0.9425). (c) Cos-1(0.4713). (d) Cos-1(- 0.4713).

(e) Tan-1(1.557). (f) Tan-1(- 1.557). (g) Sin-1(1.557). (Why the error message?)

Example. Find the angle in degrees for each of the following using the TI-92 Graphing Calculator.

(a) Sin-1(0.2337). (b) Sin-1(- 0.567). (c) Cos-1(- 0.9211). (d) Tan-1(1.743).

Solution: Since we want the angle in degrees we must use ‘DEGREE’ mode.

   2  (a)  [SIN-1] 0.2337   Displays: 13.515 (which is 13.515° )

(b)  [SIN-1]  0.567   Displays: - 34.5413 (which is - 34.5413° )

(c)  [COS-1]  0.9211   Displays: 157.087 (which is 157.087° )

(d)  [TAN-1] 1.743   Displays: 60.1561 (which is 60.1561° )

Problem. Find the angle in degrees for each of the following using the TI-92 Graphing Calculator.

(a) Sin-1(0.7833). (b) Sin-1(- 0.9425). (c) Cos-1(0.4713). (d) Cos-1(- 0.4713).

(e) Tan-1(1.557). (f) Tan-1(- 1.557). (g) Sin-1(1.557). (Why the error message?)

The TI-92 can be used to solve some trigonometric equations.

Example. Use the TI-92 calculator to solve 2 cos u = - 1, for 0 £ u < 2p .

Solution: Turn on, clear home screen and entry line:   8 

Check mode for radian. The bottom line should show RAD.

Get ‘Algebra’ menu and ‘Solve’:  1

Enter the equation: 2  u    1  u  

The display shows: 

 This answer needs some explanation and manipulation to agree with the answer when the example was done earlier. The symbol @n1 indicates an ‘arbitrary integer’ that represents any integer. In the earlier discussion, we used u ± 2p n, n = 0, 1, 2, ¼ In the arbitrary integer symbol, the number following n can be any number 1 through 255, that is @n1¼ @n255. Each time it is used the number increases. (If you clear memory, by  [MEM]  1 , the number again starts at 1. You may have to adjust the display contrast by   and  .)
Since @n1 is an arbitrary integer, let us choose it to be 0. Then u = and u = . 2p /3 is in the interval, but - 2p /3 is not. So let @n1 = 1, then u = 8p /3 and u = 4p /3. 8p /3 is not in the interval, but 4p /3 is.

Hence the answer is u = 2p /3, 4p /3

which agrees with the answer found earlier. You should check other values for @n1 to show they are outside the required interval.

Problem. Use the TI-92 calculator to solve 2 sin u = - 1, for 0 £ u < 2p .

Factoring is sometimes helpful.

Example. Solve 2 sinq cosq = sinq for 0° £ q < 360° .

Solution: Transpose sinq and factor. sinq (2 cosq - 1) = 0.

Setting each factor equal to zero and simplifying, we obtain

sinq = 0 and 2 cosq - 1 = 0

Thus sinq = 0 and cosq = 1/2

which leads to q = 0° , 180° and q = 60° , 300°

The solution may be written as q = 0° , 60° , 180° , 300°

You should check each of these values in the original equation.

Problem. Solve 2 sinq cosq = cosq for 0° £ q < 360° .

Example. Use the TI-92 to solve 2 sinq cosq = sinq for 0° £ q < 360° .

Change MODE to DEG:    2    2  [q]   [q]    [q]   [q]   The display shows: q = 60× (6× @n3 + 1) or q = 60× (6× @n3 - 1) or q = 180 × @n4 (the solid right pointing triangle ú means press B for more of the display. The line must be highlighted. If it is not, press C or D until it is.) We now choose integer values for @n3 and @n4 until we obtain all values in the interval 0° £ q < 360° . For @n3 = 0, q = 60 and - 60. For @n3 = 1, q = 420 and 300. For @n3 = 2, q = 780 and 660.

For @n4 = 0, q = 0. For @n4 = 1, q = 180. For @n4 = 2, q = 360.

 Thus in the interval 0° £ q < 360° , we have

2 = 0° , 60° , 180° , 300° .

Problem. Use the TI-92 to solve 2 sinq cosq = cosq for 0° £ q < 360° .

Example. Solve 2 sin4 u - 9 sin2 u + 4 = 0 for 0 £ u < 2p .

Solution: Factoring we obtain (2 sin2 u - 1)(sin2 u - 4) = 0

Setting each factor equal to zero and simplifying we obtain,

sin2 u = and sin2 u = 4

Upon taking square roots, we have

sin u = and sin u = ± 2

For the first expression we find

u = 

The second expression, sin u = ± 2, has no solution, as the absolute value of sin u cannot exceed 1.

Problem. Solve 2 cos4 u - 19 cos2 u + 9 = 0 for 0 £ u < 2p .

Example. Use the TI-92 to solve 2 sin4 u - 9 sin2 u + 4 = 0 for 0 £ u < 2p .

Solution: Be sure MODE is RAD.

„ 1 2c W u d d Z 4  9 c W u d d Z 2  4 Á 0 b u d ¸

The display shows: u = 71.4712 or u = 7.06858 or u = 5.49779 ú (press C and B many times to see more values) or u = 3.02699 or u = 2.35619 or u = .785398 or u = - .785398 or u = - 2.35619 or u = - 3.02699
or u = - 5.49779 or u = - 7.06858 or u = - 71.4712

For the required interval, we have

u = .785398, 2.35619, 3.02699, 5.49779

Sometimes we can specify the required interval as a domain constraint using the "with" operator, | , it is typed as 2 Í (or 2 K). We can add this notation to the entry line by pressing D and B several times until the cursor is at the right end of the entry line. Then press:

2 Í u 2 Ã Á 0  and  u 2 Â 2 2 T ¸ The display is: u = 5.49779 or u = 3.92699 or u = 2.35619 or u = .785398

Problem. Use the TI-92 to solve 2 cos4 u - 19 cos2 u + 9 = 0 for 0 £ u < 2p .

Example. Solve - 2 sin u + 3 cos u = 1 for 0 £ u < 2p .

Solution: Transposing 3 cos u and squaring both sides, we have

4 sin2 u = 1 - 6 cos u + 9 cos2 u

which simplifies to 13 cos2 u - 6 cos u - 3 = 0

Solving for cos u by the quadratic formula, we obtain

or cos u = 0.764 and cos u = - 0.302

For cos u = 0.764, and u in the required interval, u is a QI or QIV number.

To find the reference number, ur , we use a calculator.

ur = cos- 1 0.764 = 0.701

In QI, u = ur and u = 0.701.

In QIV, u = 2p - ur and u = 6.28 - 0.701 = 5.58

For cos u = - 0.302, and u in the required interval, u is a QII or QIII number.

The corresponding reference number is

ur = cos- 1½ - 0.302½ = cos- 1 0.302 = 1.26

In QII, u = p - ur and u = 3.14 - 1.26 = 1.88

In QIII, u = p + ur and u = 3.14 + 1.26 = 4.4

Since we squared both sides of the equation, we may have introduced extraneous roots. We must check them in the original equation, - 2 sin u + 3 cos u = 1.

For u = 5.58. we find - 2 sin 5.58 + 3 cos 5.58 = - 2(- 0.647) + 3(0.763) = 3.583 which is not 1.

Similarly for u = 1.88. (You should check it and the other two roots.)

Thus the required roots are u = 0.701, 4.4.

Problem. Solve 2 cos u - 3 sin u = 1 for 0 £ u < 2p .

Example. Use the TI-92 to solve - 2 sin u + 3 cos u = 1 with the domain restriction 0 £ u < 2p .

Solution: „ 1 · 2 W u d  3 X u d Á 1b ud 2 Í 0 2 Â Á u  and  u 2 Â 2 2 T ¸

The display shows u = 4.40542 or u = .701759

Problem. Use the TI-92 to solve 2 cos u - 3 sin u = 1 with the domain restriction 0 £ u < 2p .

Sometimes it is easier to solve for a multiple angle like 2q , 3q , q /2, etc., first rather than for q . When this is done, care must be taken to solve for angles in the desired interval. That is, if q is to lie between 0° and 360° , 2q must lie between 0° and 720° .

Example. Solve sinq cosq = for 0° £ q < 360° .

Solution: Replace sinq cosq by  sin 2q . Then  sin 2q =

And sin 2q = 

Since q is in the interval 0° £ q < 360° , 2q is in the interval 0° £ q < 720° .

Thus 2q = 240° , 300° , 600° , 660°

and the solutions are q = 120° , 150° , 300° , 330°

Problem. Solve sinq cosq = for 0° £ q < 360° .

Example. Solve 2x = tan x for x .

Solution: Using the TI-92 calculator, we have

„ 1 2x Á Y x d b x d ¸

The display shows: x = 92.6716 or x = 7.78988 or x = 4.60422ú (highlight this line by C and get more values by B ) or x = 0 or x = - 4.60422 or x = - 7.78988 or x = - 92.6716

 But, something is not right! From our knowledge of graphs there should be a root between 0 and 4.60422.

Because, if we let y1 = 2x and y2 = tan x and sketch their graphs we see that they intersect near x = 1.

Let’s see what happens when we put the constraint on x of 0 < x < p /2.

„ 1 2x Á Y x d b x d 2 Í 0 2  x  and  x  2 T e 2 ¸

The display shows: x = 1.16556

 This is more like it. This example shows that to use a calculator intelligently you must have a good idea of what results should be expected.

Problem. Use theTI-92 to solve 3x = tan x for x. Is there a root missing when no constraints are used?
 
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